1
00:00:08 --> 00:00:10
OK, this is quiz review day.

2
00:00:10 --> 00:00:16
The quiz coming up on Wednesday
will before this lecture the

3
00:00:16 --> 00:00:21
quiz will be this hour one
o'clock Wednesday in Walker,

4
00:00:21 --> 00:00:24
top floor of Walker,
closed book,

5
00:00:24 --> 00:00:25
all normal.

6
00:00:25 --> 00:00:31
I wrote down what we've covered
in this second part of the

7
00:00:31 --> 00:00:36
course, and actually I'm
impressed as I

8
00:00:36 --> 00:00:39
write it.
so that's chapter four on

9
00:00:39 --> 00:00:44
orthogonality and you're
remembering these -- what this

10
00:00:44 --> 00:00:48
is suggesting,
these are those columns are

11
00:00:48 --> 00:00:52
orthonormal vectors,
and then we call that matrix Q

12
00:00:52 --> 00:00:58
and the -- what's the key -- how
do we state the fact that those

13
00:00:58 --> 00:01:03
v- those columns are orthonormal
in

14
00:01:03 --> 00:01:08
terms of Q, it means that Q
transpose Q is the identity.

15
00:01:08 --> 00:01:13
So that's the matrix statement
of the -- of the property that

16
00:01:13 --> 00:01:19
the columns are orthonormal,
the dot products are either one

17
00:01:19 --> 00:01:24
or zero, and then we computed
the projections onto lines and

18
00:01:24 --> 00:01:28.56
onto
subspaces, and we used that to

19
00:01:28.56 --> 00:01:33
solve problems Ax=b in -- in the
least square sense,

20
00:01:33 --> 00:01:38
when there was no solution,
we found the best solution.

21
00:01:38 --> 00:01:42
And then finally this
Graham-Schmidt idea,

22
00:01:42 --> 00:01:48
which takes independent vectors
and lines them up,

23
00:01:48 --> 00:01:52
takes -- subtracts off the
projections of the part you've

24
00:01:52 --> 00:01:56
already done,
so that the new part is

25
00:01:56 --> 00:02:00
orthogonal and so it takes a
basis to an orthonormal basis.

26
00:02:00 --> 00:02:05.64
And you -- those calculations
involve square roots a lot

27
00:02:05.64 --> 00:02:09
because you're making things
unit vectors,

28
00:02:09 --> 00:02:11
but you should know that step.

29
00:02:11 --> 00:02:14
OK,
for determinants,

30
00:02:14 --> 00:02:18
the three big -- the big
picture is the properties of the

31
00:02:18 --> 00:02:21
determinant, one to three d-
properties one,

32
00:02:21 --> 00:02:25
two and three,
d- that define the determinant,

33
00:02:25 --> 00:02:28
and then four,
five, six through ten were

34
00:02:28 --> 00:02:29
consequences.

35
00:02:29 --> 00:02:34
Then the big formula that has n
factorial terms,

36
00:02:34 --> 00:02:38.34
half of them have plus signs
and half minus signs,

37
00:02:38.34 --> 00:02:40
and then the cofactor formula.

38
00:02:40 --> 00:02:45.03
So and which led us to a
formula for the inverse.

39
00:02:45.03 --> 00:02:48
And finally,
just so you know what's covered

40
00:02:48 --> 00:02:53
in from chapter three,
it's section six point one and

41
00:02:53 --> 00:02:59
two, so that's the basic idea of
eigenvalues and eigenvectors,

42
00:02:59 --> 00:03:02.97
the equation for the
eigenvalues, the mechanical

43
00:03:02.97 --> 00:03:07
step, this is really Ax equal
lambda x for all n eigenvectors

44
00:03:07 --> 00:03:11
at once, if we have n
independent eigenvectors,

45
00:03:11 --> 00:03:15
and then using that to compute
powers of a matrix.

46
00:03:15 --> 00:03:21
So you notice the differential
equations not on this list,

47
00:03:21 --> 00:03:26.8
because that's six point three,
that that's for the third quiz.

48
00:03:26.8 --> 00:03:27
OK.

49
00:03:27 --> 00:03:32
Shall I what I usually do for
review is to take an old exam

50
00:03:32 --> 00:03:37
and just try to pick out
questions that are significant

51
00:03:37 --> 00:03:42
and write them quickly on the
board, shall I -- shall I

52
00:03:42 --> 00:03:45
proceed that way
again?

53
00:03:45 --> 00:03:48
This -- this exam is really
old.

54
00:03:48 --> 00:03:55
November nineteen -- nineteen
eighty-four, so that was before

55
00:03:55 --> 00:03:57
the Web existed.

56
00:03:57 --> 00:04:02.29
So not only were the lectures
not on the Web,

57
00:04:02.29 --> 00:04:06
nobody even had a Web page,
my God.

58
00:04:06 --> 00:04:13
OK, so can I nevertheless
linear algebra was still as

59
00:04:13 --> 00:04:14
great as ever.

60
00:04:14 --> 00:04:19
So may I and that wasn't meant
to be a joke,

61
00:04:19 --> 00:04:23
OK, all right,
so let me just take these

62
00:04:23 --> 00:04:25
questions as they come.

63
00:04:25 --> 00:04:26
All right.

64
00:04:26 --> 00:04:27
OK.

65
00:04:27 --> 00:04:31
So the first question's about
projections.

66
00:04:31 --> 00:04:38
It says we're given the line,
the -- the vector a is the

67
00:04:38 --> 00:04:41
vector two, one,
two, and I want to find the

68
00:04:41 --> 00:04:45
projection matrix P that
projects onto the line through

69
00:04:45 --> 00:04:46
a.

70
00:04:46 --> 00:04:49.71
So my picture is,
well I'm in three dimensions,

71
00:04:49.71 --> 00:04:54
of course, so there's a vector,
two -- there's the vector a,

72
00:04:54 --> 00:04:57
two, one, two,
let me draw the whole line

73
00:04:57 --> 00:05:01
through it, and I want to
project

74
00:05:01 --> 00:05:05
any vector b onto that line,
and I'm looking for the

75
00:05:05 --> 00:05:06
projection matrix.

76
00:05:06 --> 00:05:11
So -- so this -- the projection
matrix is the matrix that I

77
00:05:11 --> 00:05:13
multiply b by to get here.

78
00:05:13 --> 00:05:18
And I guess this first part,
this is just part one a,

79
00:05:18 --> 00:05:22
I'm really asking you the --
the quick way to answer,

80
00:05:22 --> 00:05:26
to find P, is just to remember
what

81
00:05:26 --> 00:05:27
the formula is.

82
00:05:27 --> 00:05:31
And -- and we're in -- we're
projecting onto a line,

83
00:05:31 --> 00:05:35
so our formula,
our -- our usual formula is AA

84
00:05:35 --> 00:05:39
transpose A inverse A transpose,
but now A is just a column,

85
00:05:39 --> 00:05:42
one-column matrix,
so it'll be just a,

86
00:05:42 --> 00:05:46
so I'll just call it little a,
little a transpose,

87
00:05:46 --> 00:05:50
and this is just
a number now,

88
00:05:50 --> 00:05:54
one by one, so I can put it in
the denominator,

89
00:05:54 --> 00:05:59
so there's our -- that's really
what we want to remember,

90
00:05:59 --> 00:06:03
and -- and using that two,
one, two, what will I get?

91
00:06:03 --> 00:06:09
I'm dividing by -- what's the
length squared of that vector?

92
00:06:09 --> 00:06:12
So what's a transpose a?

93
00:06:12 --> 00:06:16
Looks like nine,
and what's the matrix,

94
00:06:16 --> 00:06:20.89
well, I'm -- I'm doing two,
one, two against two,

95
00:06:20.89 --> 00:06:25
one, two, so it's one-ninth of
this matrix four,

96
00:06:25 --> 00:06:28
two, four, two,
one, two, four,

97
00:06:28 --> 00:06:30
two, four.

98
00:06:30 --> 00:06:34
Now the next part asked about
eigenvalues.

99
00:06:34 --> 00:06:39
So you see we're -- since we're
learning, we know a lot more

100
00:06:39 --> 00:06:44
now, we can make connections
between these chapters,

101
00:06:44 --> 00:06:49
so what's the eigenvector,
what are the eigenvalues of P?

102
00:06:49 --> 00:06:53
I could ask what's the rank of
P.

103
00:06:53 --> 00:06:55
What's the rank of that matrix?

104
00:06:55 --> 00:06:56
Uh -- one.

105
00:06:56 --> 00:06:57
Rank is one.

106
00:06:57 --> 00:06:59
What's the column space?

107
00:06:59 --> 00:07:04.19
If I apply P to all vectors
then I fill up the column space,

108
00:07:04.19 --> 00:07:08
it's combinations of the
columns, so what's the column

109
00:07:08 --> 00:07:08
space?

110
00:07:08 --> 00:07:10
Well, it's just this line.

111
00:07:10 --> 00:07:14
The column space is the line
through two, one,

112
00:07:14 --> 00:07:14.89
two.

113
00:07:14.89 --> 00:07:18
And now what's the
eigenvalue?

114
00:07:18 --> 00:07:24
So, or since that matrix has
rank one, so tell me the

115
00:07:24 --> 00:07:27
eigenvalues of this matrix.

116
00:07:27 --> 00:07:31
It's a singular matrix,
so it certainly has an

117
00:07:31 --> 00:07:33
eigenvalue zero.

118
00:07:33 --> 00:07:40
Actually, the rank is only one,
so that means that there like

119
00:07:40 --> 00:07:44
going to be t- a
two-dimensional null space,

120
00:07:44 --> 00:07:49
there'll be at least two,
this lambda equals zero will be

121
00:07:49 --> 00:07:52
repeated twice because I can
find two independent

122
00:07:52 --> 00:07:57.08
eigenvectors with lambda equals
zero, and then of course since

123
00:07:57.08 --> 00:08:00
it's got three eigenvalues,
what's the third one?

124
00:08:00 --> 00:08:01
It's one.

125
00:08:01 --> 00:08:03
How do I know it's one?

126
00:08:03 --> 00:08:08
Either from the trace,
which is nine over nine,

127
00:08:08 --> 00:08:12
which is one,
or by remembering what -- what

128
00:08:12 --> 00:08:16
is the eigenvector,
and actually now it's going to

129
00:08:16 --> 00:08:21
ask for the eigenvector,
so what's the eigenvector for

130
00:08:21 --> 00:08:23
that eigenvalue?

131
00:08:23 --> 00:08:28
What's the eigenvector for that
eigenvalue?

132
00:08:28 --> 00:08:32
It's the -- it's the vector
that doesn't move,

133
00:08:32 --> 00:08:37
eigenvalue one,
so the vector that doesn't move

134
00:08:37 --> 00:08:37
is a.

135
00:08:37 --> 00:08:43
This a is the -- is -- is also
the eigenvector with lambda

136
00:08:43 --> 00:08:48
equal one, because if I apply
the projection matrix to a,

137
00:08:48 --> 00:08:51
I get a again.

138
00:08:51 --> 00:08:56
Everybody sees that if I apply
that matrix to a,

139
00:08:56 --> 00:09:02.21
can I do it in little letters,
if I apply that matrix to a,

140
00:09:02.21 --> 00:09:07
then I have a transpose a
canceling a transpose a and I

141
00:09:07 --> 00:09:09
get a again.

142
00:09:09 --> 00:09:12
So sure enough,
Pa equals a.

143
00:09:12 --> 00:09:14
And the eigenvalue is one.

144
00:09:14 --> 00:09:15
OK.

145
00:09:15 --> 00:09:18
Good.
now, actually it asks you

146
00:09:18 --> 00:09:25.58
further to solve
this difference equation,

147
00:09:25.58 --> 00:09:34
so this will be -- this is --
this is solve u(k+1)=Puk,

148
00:09:34 --> 00:09:40
starting from u0 equal nine,
nine, zero.

149
00:09:40 --> 00:09:42
And find uk.

150
00:09:42 --> 00:09:45
So -- so what's up?

151
00:09:45 --> 00:09:50
Shall we find u1 first of all?

152
00:09:50 --> 00:09:53
So just to get started.

153
00:09:53 --> 00:09:55
So what is u1?

154
00:09:55 --> 00:10:00
It's Pu0 of course.

155
00:10:00 --> 00:10:06
So if I do the projection of --
of this vector onto the line,

156
00:10:06 --> 00:10:12
so this is like my vector b now
that I'm projecting onto the

157
00:10:12 --> 00:10:17
line, I get a times a transpose
u0 over a transpose a.

158
00:10:17 --> 00:10:22
Well, one way or another I just
do this multiplication.

159
00:10:22 --> 00:10:27
but maybe this is the easiest
way to do it.

160
00:10:27 --> 00:10:31
a transpose,
can I remember what a is on

161
00:10:31 --> 00:10:31
this board?

162
00:10:31 --> 00:10:35.49
Two, one, two,
so I'm projecting onto the line

163
00:10:35.49 --> 00:10:36
through there.

164
00:10:36 --> 00:10:40
This is the projection,
it's P times the vector u0,

165
00:10:40 --> 00:10:44
so what do I have for a
transpose u0 looks like

166
00:10:44 --> 00:10:46
eighteen, looks like
twenty-seven,

167
00:10:46 --> 00:10:50
and a transpose a we figured
was nine,

168
00:10:50 --> 00:10:55
so it's three a,
so that this is the -- this is

169
00:10:55 --> 00:11:02
the x hat, the -- the multiple
of a, in -- in our formulas,

170
00:11:02 --> 00:11:06
and of course that's six,
three, six.

171
00:11:06 --> 00:11:07
So that's u1.

172
00:11:07 --> 00:11:09
Computed out directly.

173
00:11:09 --> 00:11:16
That's on the line through a
and it's the closest point to

174
00:11:16 --> 00:11:18
u0, and it's just Pu0.

175
00:11:18 --> 00:11:23
You just
straightforward multiplication

176
00:11:23 --> 00:11:25
produces that.

177
00:11:25 --> 00:11:25
OK.

178
00:11:25 --> 00:11:27
Now, what's u2?

179
00:11:27 --> 00:11:29.97
Well, u2 is Pu1,
I agree.

180
00:11:29.97 --> 00:11:33
Do I need to compute that
again?

181
00:11:33 --> 00:11:38
No, because once I'm already on
the line through A,

182
00:11:38 --> 00:11:43
uk will be, I could do the
projection k times,

183
00:11:43 --> 00:11:47
but it's enough just to do it
once.

184
00:11:47 --> 00:11:51
It's the same,
it's the same,

185
00:11:51 --> 00:11:53
six, three, six.

186
00:11:53 --> 00:11:59
So this is a case where I could
and -- and actually on the quiz

187
00:11:59 --> 00:12:04
if you see one of these,
which could very well be there,

188
00:12:04 --> 00:12:08
and it could very well be not a
projection matrix,

189
00:12:08 --> 00:12:13
then we would use all the
eigenvalues and eigenvectors.

190
00:12:13 --> 00:12:17
Let's think for a moment,
how do

191
00:12:17 --> 00:12:18
you do those?

192
00:12:18 --> 00:12:23
M - the point of this small
part of a question was that when

193
00:12:23 --> 00:12:27
P is a projection matrix,
so that P squared equals P and

194
00:12:27 --> 00:12:30
P cubed equals P,
then -- then we don't need to

195
00:12:30 --> 00:12:34
get into the mechanics of all
knowing all the other

196
00:12:34 --> 00:12:37
eigenvalues and eigenvectors.

197
00:12:37 --> 00:12:40
We just
can go directly.

198
00:12:40 --> 00:12:45
But if P was now some other
matrix, can you just -- let's

199
00:12:45 --> 00:12:51
just remember from these very
recent lectures how you would

200
00:12:51 --> 00:12:55
proceed.
from these very recent lectures

201
00:12:55 --> 00:13:01
we know that uk we would -- we
would expand u0 as a combination

202
00:13:01 --> 00:13:03
of eigenvectors.

203
00:13:03 --> 00:13:07
Let me leave -- yeah,
as a combination of

204
00:13:07 --> 00:13:12
eigenvectors,
c1x1, some multiple of the

205
00:13:12 --> 00:13:17
second eigenvector,
some multiple of the third

206
00:13:17 --> 00:13:21
eigenvector, and then A to the
ku0 would be c1,

207
00:13:21 --> 00:13:26
so this -- we have to find
these numbers here,

208
00:13:26 --> 00:13:29
that's the work actually.

209
00:13:29 --> 00:13:33
The work
is find the eigenvalues,

210
00:13:33 --> 00:13:39
find the eigenvectors and find
the c-s because they all come

211
00:13:39 --> 00:13:40
into the formula.

212
00:13:40 --> 00:13:46
We have -- so -- so to do this,
you can see what you have to

213
00:13:46 --> 00:13:47
compute.

214
00:13:47 --> 00:13:52
You have to compute the
eigenvalues, you have to compute

215
00:13:52 --> 00:13:55
the eigenvectors,
and then to match u0 you

216
00:13:55 --> 00:13:58
compute the c-s,
and

217
00:13:58 --> 00:14:00
then you've got it.

218
00:14:00 --> 00:14:05
So it's -- it's just that's a
formula that shows what pieces

219
00:14:05 --> 00:14:06
we need.

220
00:14:06 --> 00:14:11
And what would actually happen
in the case of this projection

221
00:14:11 --> 00:14:12
matrix?

222
00:14:12 --> 00:14:16
If this A is a projection
matrix, then a couple of

223
00:14:16 --> 00:14:18
eigenvalues are zero.

224
00:14:18 --> 00:14:22
That's why we just throw those
away.

225
00:14:22 --> 00:14:27
The other eigenvalue was a one,
so that we got the same thing

226
00:14:27 --> 00:14:29
every time, c3x3.

227
00:14:29 --> 00:14:32
From the first time,
second time,

228
00:14:32 --> 00:14:36
third, all iterations pro- left
us with this constant,

229
00:14:36 --> 00:14:40
left us right here at six,
three, six.

230
00:14:40 --> 00:14:46
But maybe I take -- I'm taking
this chance to remind you

231
00:14:46 --> 00:14:49
of what to do for other
matrices.

232
00:14:49 --> 00:14:49
OK.

233
00:14:49 --> 00:14:51
So that's part way through.

234
00:14:51 --> 00:14:51
OK.

235
00:14:51 --> 00:14:56
The next question in nineteen
eighty-four is fitting a

236
00:14:56 --> 00:14:58.51
straight line to points.

237
00:14:58.51 --> 00:15:02
And actually a straight line
through the origin.

238
00:15:02 --> 00:15:05
A straight line through the
origin.

239
00:15:05 --> 00:15:07.91
So can I go to question two?

240
00:15:07.91 --> 00:15:12
So this is fitting a straight
line to

241
00:15:12 --> 00:15:19
these points,
can I -- I'll just give you the

242
00:15:19 --> 00:15:26
points at t=1 the y is four,
at t=2, y is five,

243
00:15:26 --> 00:15:29
at t=3, y is eight.

244
00:15:29 --> 00:15:36
So we've got points one,
two, three, four,

245
00:15:36 --> 00:15:39
five, and eight.

246
00:15:39 --> 00:15:47
And I'm trying to fit a
straight line through the origin

247
00:15:47 --> 00:15:52
to
these three values.

248
00:15:52 --> 00:15:57
OK, so my equation that I'm
allowing myself is just y equal

249
00:15:57 --> 00:15:58
Dt.

250
00:15:58 --> 00:16:01
So I have only one unknown.

251
00:16:01 --> 00:16:03
One degree of freedom.

252
00:16:03 --> 00:16:04
One parameter D.

253
00:16:04 --> 00:16:11
So I'm expecting to end up my
matrix so my -- my -- when I try

254
00:16:11 --> 00:16:16
to -- when I try to fit a
straight line,

255
00:16:16 --> 00:16:21
that goes through the origin,
that's because it goes through

256
00:16:21 --> 00:16:24
the origin, I've lost the
constant c here,

257
00:16:24 --> 00:16:28
so I have just this should be a
quick calculation.

258
00:16:28 --> 00:16:32
and I can write down the three
equations that -- that -- that

259
00:16:32 --> 00:16:37.27
would -- I'd like to solve if
the line went through the

260
00:16:37.27 --> 00:16:40
points, that's a good start.

261
00:16:40 --> 00:16:43
Because that displays the
matrix.

262
00:16:43 --> 00:16:46
So can I continue that problem?

263
00:16:46 --> 00:16:52
We would like to solve-- so y
is Dt, so I'd like to solve D

264
00:16:52 --> 00:16:57
times one times D equals four,
two times D equals five and

265
00:16:57 --> 00:17:00
three times D equals eight.

266
00:17:00 --> 00:17:02
That would be perfection.

267
00:17:02 --> 00:17:09
If I could find such a D,
then the line y equal Dt would

268
00:17:09 --> 00:17:14
satisfy all three equations,
would go through all three

269
00:17:14 --> 00:17:17
points, but it doesn't exist.

270
00:17:17 --> 00:17:22
So -- so I have to solve this
-- so the -- my matrix is now

271
00:17:22 --> 00:17:26.73
you can see my matrix,
it just has one column.

272
00:17:26.73 --> 00:17:28.89
Multiplying a scalar D.

273
00:17:28.89 --> 00:17:32
And you can see the right-hand
side.

274
00:17:32 --> 00:17:34
This is my Ax=b.

275
00:17:34 --> 00:17:39
I don't need three equals signs
now because I've got vectors.

276
00:17:39 --> 00:17:39
OK.

277
00:17:39 --> 00:17:43
There's Ax=b and you take it
from there.

278
00:17:43 --> 00:17:48
You the -- the best x will be
-- will come from -- so what's

279
00:17:48 --> 00:17:50
the -- the key equation?

280
00:17:50 --> 00:17:53
So this is the A,
this is the Ax hat equal b

281
00:17:53 --> 00:17:54
equation.

282
00:17:54 --> 00:17:55
Well, Ax=b.

283
00:17:55 --> 00:17:59
And what's the equation
for x hat?

284
00:17:59 --> 00:18:05
The best D, so to find the best
D, the best x,

285
00:18:05 --> 00:18:12
the equation is A transpose A,
the best D, is A transpose

286
00:18:12 --> 00:18:16
times the right-hand side.

287
00:18:16 --> 00:18:21
This is all coming from
projection on a line,

288
00:18:21 --> 00:18:27.73
our -- our matrix only has one
column.

289
00:18:27.73 --> 00:18:32
So A transpose A would be maybe
fourteen, D hat,

290
00:18:32 --> 00:18:38
and A transpose b I'm getting
four, ten, and twenty-four.

291
00:18:38 --> 00:18:40
Is that right?

292
00:18:40 --> 00:18:42
Four, ten and twenty-four.

293
00:18:42 --> 00:18:44
So thirty-eight.

294
00:18:44 --> 00:18:50.9
So that tells me the best D hat
is D hat is thirty-eight over

295
00:18:50.9 --> 00:18:53
fourteen.

296
00:18:53 --> 00:18:53
OK.

297
00:18:53 --> 00:18:53.73
Fine.

298
00:18:53.73 --> 00:18:56
All right.
so we found the best line.

299
00:18:56 --> 00:19:00
And now here's a -- here's the
next question.

300
00:19:00 --> 00:19:04
What vector did I just project
onto what line?

301
00:19:04 --> 00:19:09
See in this section on least
squares here's the key point,

302
00:19:09 --> 00:19:15
I'm -- I'm asking you to think
of the least squares problem in

303
00:19:15 --> 00:19:17
two
ways.

304
00:19:17 --> 00:19:18
Two different pictures.

305
00:19:18 --> 00:19:20
Two different graphs.

306
00:19:20 --> 00:19:21
One graph is this.

307
00:19:21 --> 00:19:25.78
This is a graph in the -- in
the b -- in the tb plane,

308
00:19:25.78 --> 00:19:26.45
ty plane.

309
00:19:26.45 --> 00:19:28
The -- the -- the line itself.

310
00:19:28 --> 00:19:32
The other picture I'm asking
you to think of is like my

311
00:19:32 --> 00:19:34
projection picture.

312
00:19:34 --> 00:19:39.44
What -- what projection --
what -- what vector I -- I

313
00:19:39.44 --> 00:19:45
projecting onto what line or
what subspace when I -- when I

314
00:19:45 --> 00:19:45
do this?

315
00:19:45 --> 00:19:51
So the -- my second picture is
a projection picture that --

316
00:19:51 --> 00:19:55
that sees the whole thing with
vectors.

317
00:19:55 --> 00:19:59
Here's my vector of course that
I'm projecting.

318
00:19:59 --> 00:20:05
I'm projecting that vector b
onto the column space of A.

319
00:20:05 --> 00:20:11
Of if you like -- it's just a
line onto that's the line it's

320
00:20:11 --> 00:20:14
just a line, of course.

321
00:20:14 --> 00:20:18
That's what this calculation is
doing.

322
00:20:18 --> 00:20:24
This is computing the best D,
which is -- this is the x hat.

323
00:20:24 --> 00:20:29
So -- so seeing it as a
projection means I don't see the

324
00:20:29 --> 00:20:33.08
projection in this figure,
right?

325
00:20:33.08 --> 00:20:37
In this figure I'm not
projecting

326
00:20:37 --> 00:20:43
those points onto that line or
anything of the sort.

327
00:20:43 --> 00:20:49
The projection s-picture for --
for least squares is in the --

328
00:20:49 --> 00:20:54
in the space where b lies,
the whole vector b,

329
00:20:54 --> 00:20:56
and the columns of A.

330
00:20:56 --> 00:21:02
And then the x is the best
combination that gives the

331
00:21:02 --> 00:21:03
projection.

332
00:21:03 --> 00:21:03
OK.

333
00:21:03 --> 00:21:08
So that's a chance to tell me
that.

334
00:21:08 --> 00:21:08
OK.

335
00:21:08 --> 00:21:15
I'll go -- OK now finally in
orthogonality there's the

336
00:21:15 --> 00:21:17
Graham-Schmidt idea.

337
00:21:17 --> 00:21:20
So that's problem two D here.

338
00:21:20 --> 00:21:26
It asks me if I have two
vectors, a1 equal one,

339
00:21:26 --> 00:21:33
two, three, and a2 equal one,
one, one, find two orthogonal

340
00:21:33 --> 00:21:37
vectors in that plane.

341
00:21:37 --> 00:21:42
So those two vectors give a
plane, they give a plane.

342
00:21:42 --> 00:21:48
Which is of course the -- the
column space of the -- of the

343
00:21:48 --> 00:21:49
matrix.

344
00:21:49 --> 00:21:54
And I'm looking for an
orthogonal basis for that plane.

345
00:21:54 --> 00:21:59
So I'm looking for two
orthogonal vectors.

346
00:21:59 --> 00:22:03
And of course there are lots of
--

347
00:22:03 --> 00:22:06
I mean, I've got a plane there.

348
00:22:06 --> 00:22:09
If I get one orthogonal pair,
I can rotate it.

349
00:22:09 --> 00:22:12
There's not just one answer
here.

350
00:22:12 --> 00:22:16
But Graham-Schmidt says OK,
start with the first vector,

351
00:22:16 --> 00:22:19
and let that be -- and keep
that one.

352
00:22:19 --> 00:22:23
And then take the second one
orthogonal to this.

353
00:22:23 --> 00:22:27
So -- so Graham-Schmidt says
start

354
00:22:27 --> 00:22:31
with this one and then make a
second vector B,

355
00:22:31 --> 00:22:35
can I call that second vector
B, this is going to be

356
00:22:35 --> 00:22:38
orthogonal to,
so perpendicular to a1.

357
00:22:38 --> 00:22:42
If I can with my chalk create
the key equation.

358
00:22:42 --> 00:22:46
This vector B is going to be
this one, one,

359
00:22:46 --> 00:22:49
one, but that one,
one --

360
00:22:49 --> 00:22:53
one, one, one is not
perpendicular to a1,

361
00:22:53 --> 00:22:59
so I have to subtract off its
projection, I have to subtract

362
00:22:59 --> 00:23:04
off the B, the -- the B trans-
ye the B -- the -- the I should

363
00:23:04 --> 00:23:08
say the a1 transpose b over a1
transpose a1,

364
00:23:08 --> 00:23:12
that multiple of a1,
I've got to remove.

365
00:23:12 --> 00:23:17
So I just have
to compute what that is,

366
00:23:17 --> 00:23:22
and I get ano- I get a vector B
that's orthogonal to a1.

367
00:23:22 --> 00:23:28
It's the -- it's -- it's the
original vector minus its

368
00:23:28 --> 00:23:29
projection.

369
00:23:29 --> 00:23:33.3
Oh, so what is -- I mean this
to be a2.

370
00:23:33.3 --> 00:23:33
Yeah.

371
00:23:33 --> 00:23:38
So I'm projecting a2 onto the
line through a1.

372
00:23:38 --> 00:23:42.97
That's the part that I don't
want

373
00:23:42.97 --> 00:23:46
because that's in the direction
I already have,

374
00:23:46 --> 00:23:50.62
so I subtract off that
projection and I get the part I

375
00:23:50.62 --> 00:23:52.63
want, the orthogonal part.

376
00:23:52.63 --> 00:23:52
OK.

377
00:23:52 --> 00:23:57
So that's the Graham-Schmidt
thing and we can put numbers in.

378
00:23:57 --> 00:24:01
one, one, one take away a1
transpose a2 is six,

379
00:24:01 --> 00:24:06
a1 transpose a1 is
fourteen,multiplying a1.

380
00:24:06 --> 00:24:11
And that gives us the new
orthogonal vector B.

381
00:24:11 --> 00:24:16
Because I only ask for
orthogonal right now,

382
00:24:16 --> 00:24:23
I don't have to divide by the
length which will involve a

383
00:24:23 --> 00:24:24
square root.

384
00:24:24 --> 00:24:25
OK.

385
00:24:25 --> 00:24:26
Third question.

386
00:24:26 --> 00:24:28
Third question.

387
00:24:28 --> 00:24:34.15
All right,
let me -- I'll move this board

388
00:24:34.15 --> 00:24:37
up.
third question will probably be

389
00:24:37 --> 00:24:39
about eigenvalues.

390
00:24:39 --> 00:24:39
OK.

391
00:24:39 --> 00:24:40.17
Three.

392
00:24:40.17 --> 00:24:43.13
This is a four-by-four matrix.

393
00:24:43.13 --> 00:24:48
Its eigenvalues are lambda one,
lambda two, lambda three,

394
00:24:48 --> 00:24:49
lambda four.

395
00:24:49 --> 00:24:51
Question one.

396
00:24:51 --> 00:24:56.65
What's the condition on the
lambdas so that the matrix is

397
00:24:56.65 --> 00:24:57
invertible?

398
00:24:57 --> 00:24:58
OK.

399
00:24:58 --> 00:25:05.53
So under what conditions on the
lambdas will the matrix be

400
00:25:05.53 --> 00:25:06
invertible?

401
00:25:06 --> 00:25:08
So that's easy.

402
00:25:08 --> 00:25:13
Invertible if what's the
condition on the lambdas?

403
00:25:13 --> 00:25:16
None of them are zero.

404
00:25:16 --> 00:25:22
A zero eigenvalue would mean
something in the null space

405
00:25:22 --> 00:25:27
would mean a solution to Ax=0x,
but we're invertible,

406
00:25:27 --> 00:25:31
so none of
them is zero,

407
00:25:31 --> 00:25:37
the product -- however you want
to say, no -- no zero

408
00:25:37 --> 00:25:38
eigenvalues.

409
00:25:38 --> 00:25:38
Good.

410
00:25:38 --> 00:25:42.96
OK, what's the determinant of A
inverse?

411
00:25:42.96 --> 00:25:45
The determinant of A inverse?

412
00:25:45 --> 00:25:49
So where is that going to come
from?

413
00:25:49 --> 00:25:55
Well, if we knew the
eigenvalues of A inverse,

414
00:25:55 --> 00:26:00.13
we could multiply them together
to find the determinant.

415
00:26:00.13 --> 00:26:03
And we do know the eigenvalues
of A inverse.

416
00:26:03 --> 00:26:04
What are they?

417
00:26:04 --> 00:26:09
They're just one over lambda
one times one over lambda two,

418
00:26:09 --> 00:26:14
that's the second eigenvalue,
the third eigenvalue and the

419
00:26:14 --> 00:26:14
fourth.

420
00:26:14 --> 00:26:20
So the product of the four
eigenvalues of the inverse

421
00:26:20 --> 00:26:25
will give us the determinant of
the inverse.

422
00:26:25 --> 00:26:25
Fine.

423
00:26:25 --> 00:26:26
OK.

424
00:26:26 --> 00:26:30
And what's the trace of A plus
I?

425
00:26:30 --> 00:26:34
So what do we know about trace?

426
00:26:34 --> 00:26:41.08
It's the sum down the diagonal,
but we don't know what our

427
00:26:41.08 --> 00:26:42
matrix is.

428
00:26:42 --> 00:26:47
The trace is also the sum of
the eigenvalues,

429
00:26:47 --> 00:26:53.65
and we do know the eigenvalues
of A

430
00:26:53.65 --> 00:26:54
plus I.

431
00:26:54 --> 00:26:56
So we just add them up.

432
00:26:56 --> 00:27:01.39
So what -- what's the first
eigenvalue of A plus I?

433
00:27:01.39 --> 00:27:05
When the matrix A has
eigenvalues lambda one,

434
00:27:05 --> 00:27:10
two, three and four,
then the eigenvalues if I add

435
00:27:10 --> 00:27:14
the identity,
that moves all the eigenvalues

436
00:27:14 --> 00:27:18
by one,
so I just add up lambda one

437
00:27:18 --> 00:27:23
plus one, lambda two plus one,
and so on, lambda three plus

438
00:27:23 --> 00:27:28
one, lambda four plus one,
so it's lambda one plus lambda

439
00:27:28 --> 00:27:32
two plus lambda three plus
lambda four plus four.

440
00:27:32 --> 00:27:33
Right.

441
00:27:33 --> 00:27:38
That movement by the identity
moved all the eigenvalues by

442
00:27:38 --> 00:27:43
one, so it moved
the whole trace by four.

443
00:27:43 --> 00:27:47
So it was the trace of A plus
four more.

444
00:27:47 --> 00:27:47
OK.

445
00:27:47 --> 00:27:48
Let's see.

446
00:27:48 --> 00:27:53
We may be finished this quiz
twenty minutes early.

447
00:27:53 --> 00:27:53
No.

448
00:27:53 --> 00:27:56
There's another question.

449
00:27:56 --> 00:27:57
Oh, God, OK.

450
00:27:57 --> 00:28:00
How did this class ever do it?

451
00:28:00 --> 00:28:05
Well, you'll see.
you'll be able to do it.

452
00:28:05 --> 00:28:08
OK.
this has got to be a

453
00:28:08 --> 00:28:10
determinant question.

454
00:28:10 --> 00:28:11
All right.

455
00:28:11 --> 00:28:17
More determinants and cofactors
and big formula question.

456
00:28:17 --> 00:28:17
OK.

457
00:28:17 --> 00:28:19
Let me do that.

458
00:28:19 --> 00:28:23
So it's about a matrix,
a -- a whole family of

459
00:28:23 --> 00:28:24
matrices.

460
00:28:24 --> 00:28:27
Here's the four-by-four one.

461
00:28:27 --> 00:28:34
The four-by-four one is,
and -- and all the matrices in

462
00:28:34 --> 00:28:38
this family are tridiagonal with
-- with ones.

463
00:28:38 --> 00:28:39
Otherwise zeroes.

464
00:28:39 --> 00:28:44
So that's the pattern,
and we've seen this matrix.

465
00:28:44 --> 00:28:44.48
OK.

466
00:28:44.48 --> 00:28:49
So the -- it's tridiagonal with
ones on the diagonal,

467
00:28:49 --> 00:28:54
ones above and ones below,
and you see the general formula

468
00:28:54 --> 00:28:58.56
An, so I'll use Dn for the
determinant

469
00:28:58.56 --> 00:28:59
of An.

470
00:28:59 --> 00:28:59
OK.

471
00:28:59 --> 00:29:01
All right.

472
00:29:01 --> 00:29:09
So I'm going to do a -- the
first question is use cofactors

473
00:29:09 --> 00:29:16
to show that Dn is something
times D(n-1) plus something

474
00:29:16 --> 00:29:18
times D(n-2).

475
00:29:18 --> 00:29:22
And find those somethings.

476
00:29:22 --> 00:29:22
OK.

477
00:29:22 --> 00:29:30
So this -- the fact that it's
tridiagonal with

478
00:29:30 --> 00:29:35
these constant diagonals means
that there is such a recurrence

479
00:29:35 --> 00:29:36.62
formula.

480
00:29:36.62 --> 00:29:39
And so the first question is
find it.

481
00:29:39 --> 00:29:42
Well, what's the recurrence
formula?

482
00:29:42 --> 00:29:44
OK, how does it go?

483
00:29:44 --> 00:29:47
So I'll use cofactors along the
first row.

484
00:29:47 --> 00:29:52
So I take that number times its
cofactor.

485
00:29:52 --> 00:29:57
So it's one times its cofactor
and what is its cofactor?

486
00:29:57 --> 00:30:01
D(n-1), right,
exactly, the cofactor is this

487
00:30:01 --> 00:30:05.15
-- is this guy uses up row one
and column one,

488
00:30:05.15 --> 00:30:09
so the cofactor is down here,
so it's one of those.

489
00:30:09 --> 00:30:13
OK, that's the first cofactor
term.

490
00:30:13 --> 00:30:17
Now the other cofactor term is
this guy.

491
00:30:17 --> 00:30:23
Which uses up row one and
column two and what's surprising

492
00:30:23 --> 00:30:24
about that?

493
00:30:24 --> 00:30:29.93
When you use row one and column
two that brings in a minus.

494
00:30:29.93 --> 00:30:35
There'll be a minus because the
-- the cofactor is this

495
00:30:35 --> 00:30:37.97
determinant times minus one.

496
00:30:37.97 --> 00:30:43
The the one-two cofactor
is that determinant with its

497
00:30:43 --> 00:30:44
sign changed.

498
00:30:44 --> 00:30:45
OK.

499
00:30:45 --> 00:30:49
So I have to look at that
determinant and I have to

500
00:30:49 --> 00:30:53
remember in my head a sign is
going to get changed.

501
00:30:53 --> 00:30:53
OK.

502
00:30:53 --> 00:30:56
Now how do I do that
determinant?

503
00:30:56 --> 00:30:58
How do I make that one clear?

504
00:30:58 --> 00:31:02.53
I -- the --
the neat way to do is -- is

505
00:31:02.53 --> 00:31:06
here I see I -- I'll use
cofactors down the first column.

506
00:31:06 --> 00:31:10
Because the first column is all
zeroes except for that one,

507
00:31:10 --> 00:31:13
so this one is now -- and
what's its cofactor?

508
00:31:13 --> 00:31:18
Within this three-by-three its
cofactor will be two-by-two,

509
00:31:18 --> 00:31:19
and what is it?

510
00:31:19 --> 00:31:21
It's this, right?

511
00:31:21 --> 00:31:26
So -- so that part is all gone,
so I'm taking that times its

512
00:31:26 --> 00:31:31
cofactor, then zero times
whatever its cofactor is,

513
00:31:31 --> 00:31:36
so it's really just one times
and what's this in the general

514
00:31:36 --> 00:31:37
n-by-n case?

515
00:31:37 --> 00:31:40
It's Dn minus two.

516
00:31:40 --> 00:31:43
But now so is this a plus or
sign or a minus sign,

517
00:31:43 --> 00:31:47
it's -- it's just a one,
because there's a one from

518
00:31:47 --> 00:31:49
there and a one from there.

519
00:31:49 --> 00:31:51.31
And is it a plus or a minus?

520
00:31:51.31 --> 00:31:55
It's minus I guess because
there was a minus the first time

521
00:31:55 --> 00:31:59
and then the second time
it's a plus,

522
00:31:59 --> 00:32:02.27
so it's overall it's a minus.

523
00:32:02.27 --> 00:32:06.93
So there's my a and b were one
and minus one.

524
00:32:06.93 --> 00:32:08
Those constants.

525
00:32:08 --> 00:32:12
Th- that's the -- that's the
recurrence.

526
00:32:12 --> 00:32:13
OK.

527
00:32:13 --> 00:32:19
And oh, then it asks you to
then it asks you to solve this

528
00:32:19 --> 00:32:24
thing first by
writing it as a -- as a system.

529
00:32:24 --> 00:32:27
So now I'd like to know the
solution.

530
00:32:27 --> 00:32:31
I -- I better know how it
starts, right?

531
00:32:31 --> 00:32:34.28
It starts with D1,
what was D1,

532
00:32:34.28 --> 00:32:38
that's just the one-by-one
case, so D1 is one,

533
00:32:38 --> 00:32:40
and what is D2?

534
00:32:40 --> 00:32:44
Just to get us started and then
this would give us D3,

535
00:32:44 --> 00:32:46
D4, and forever.

536
00:32:46 --> 00:32:50
D2 is this two-by-two that I'm
seeing here and that determinant

537
00:32:50 --> 00:32:52
is obviously zero.

538
00:32:52 --> 00:32:57
So those little ones will start
the recurrence and then we take

539
00:32:57 --> 00:32:57
off.

540
00:32:57 --> 00:33:02.32
And then the idea is to write
this recurrence as --

541
00:33:02.32 --> 00:33:09
as a Dn, D(n-1) is some matrix
times the one before,

542
00:33:09 --> 00:33:12
the D(n-1), D(n-2).

543
00:33:12 --> 00:33:14
What's the matrix?

544
00:33:14 --> 00:33:23
You see, you remember this step
of taking a single second order

545
00:33:23 --> 00:33:31.75
equation and by introducing a
vector unknown to make it into a

546
00:33:31.75 --> 00:33:35
--
to a first order system.

547
00:33:35 --> 00:33:35
OK.

548
00:33:35 --> 00:33:41.6
So Dn is one of Dn minus one
minus one, I think that -- that

549
00:33:41.6 --> 00:33:44
goes in the first row,
right?

550
00:33:44 --> 00:33:46
From the equation above?

551
00:33:46 --> 00:33:51
And the second one is this is
the same as this,

552
00:33:51 --> 00:33:53
so one and zero are fine.

553
00:33:53 --> 00:33:55
So there's the matrix.

554
00:33:55 --> 00:33:57
OK.

555
00:33:57 --> 00:34:00
So now how do I proceed?

556
00:34:00 --> 00:34:06
We can guess what this
examiner's got in his little

557
00:34:06 --> 00:34:10
mind.
well, find the eigenvalues.

558
00:34:10 --> 00:34:17
And actually it tells us that
the sixth power of these

559
00:34:17 --> 00:34:21
eigenvalues turns out to be one.

560
00:34:21 --> 00:34:30
Uh, well, can -- can we get the
equation for the eigenvalues?

561
00:34:30 --> 00:34:34
Let's do it and let's get a
formula for them.

562
00:34:34 --> 00:34:34
OK.

563
00:34:34 --> 00:34:36
So what are the eigenvalues?

564
00:34:36 --> 00:34:40
I look at the -- the matrix,
this determinant one minus

565
00:34:40 --> 00:34:45.31
lambda and zero minus lambda,
and these guys are still there,

566
00:34:45.31 --> 00:34:49
I compute that determinant,
I get lambda squared minus

567
00:34:49 --> 00:34:52
lambda
and then plus one.

568
00:34:52 --> 00:34:54
And I set that to zero.

569
00:34:54 --> 00:34:54
OK.

570
00:34:54 --> 00:34:57
So we're not Fibonacci here.

571
00:34:57 --> 00:35:01
We're -- we're not seeing
Fibonacci numbers.

572
00:35:01 --> 00:35:06
Because the sign -- we had a
sign change there.

573
00:35:06 --> 00:35:12.25
And it's not clear right away
whether these -- whether this --

574
00:35:12.25 --> 00:35:14.49
is it clear?

575
00:35:14.49 --> 00:35:17
Is this matrix stable or
unstable?

576
00:35:17 --> 00:35:21
When we take -- when we go
further and further out?

577
00:35:21 --> 00:35:23
Are these Ds increasing?

578
00:35:23 --> 00:35:25
Are they going to zero?

579
00:35:25 --> 00:35:28
Are they bouncing around
periodically?

580
00:35:28 --> 00:35:30
the answers have to be here.

581
00:35:30 --> 00:35:34
I would like to know how big
these lambdas are,

582
00:35:34 --> 00:35:35
right?

583
00:35:35 --> 00:35:41
And the point is probably these
-- let's -- let's see,

584
00:35:41 --> 00:35:42
what's lambda?

585
00:35:42 --> 00:35:47
From the quadratic formula
lambda is one,

586
00:35:47 --> 00:35:53
I switch the sign of that,
plus or minus the square root

587
00:35:53 --> 00:35:58
of one minus 4ac,
I getting a minus three there?

588
00:35:58 --> 00:35:59
Over two.

589
00:35:59 --> 00:36:00
What's up?

590
00:36:00 --> 00:36:03
They're complex.

591
00:36:03 --> 00:36:10
The -- the eigenvalues are one
plus square root of three I over

592
00:36:10 --> 00:36:15
two and one minus square root of
three I over two.

593
00:36:15 --> 00:36:18
What's the magnitude of lambda?

594
00:36:18 --> 00:36:22
That's the key point for
stability.

595
00:36:22 --> 00:36:26
These are two numbers in the
complex plane.

596
00:36:26 --> 00:36:30
One plus some --
somewhere here,

597
00:36:30 --> 00:36:33
and its complex conjugate
there.

598
00:36:33 --> 00:36:37
I want to know how far from the
origin are those numbers.

599
00:36:37 --> 00:36:40
What's the magnitude of lambda?

600
00:36:40 --> 00:36:42
And do you see what it is?

601
00:36:42 --> 00:36:46
Do you recognize this -- a
number like that?

602
00:36:46 --> 00:36:51
Take the real part squared and
the imaginary part squared and

603
00:36:51 --> 00:36:52
add.

604
00:36:52 --> 00:36:53
What do you get?

605
00:36:53 --> 00:36:56
So the real part squared is a
quarter.

606
00:36:56 --> 00:37:00
The imaginary part squared is
three-quarters.

607
00:37:00 --> 00:37:01
They add to one.

608
00:37:01 --> 00:37:06
That's a number with -- that's
on the unit circle.

609
00:37:06 --> 00:37:08
That's an e to the i theta.

610
00:37:08 --> 00:37:11
That's a
cos(theta)+isin(theta).

611
00:37:11 --> 00:37:13
And what's theta?

612
00:37:13 --> 00:37:19
This -- this is like a complex
number that's worth knowing,

613
00:37:19 --> 00:37:22
it's not totally obvious but
it's nice.

614
00:37:22 --> 00:37:27
That's -- I should see that as
cos(theta)+isin(theta),

615
00:37:27 --> 00:37:31
and the angle that would do
that is

616
00:37:31 --> 00:37:33
sixty degrees,
pi over three.

617
00:37:33 --> 00:37:37
So that's a -- let me improve
my picture.

618
00:37:37 --> 00:37:42
So those -- that's e to the i
pi over six -- pi over three.

619
00:37:42 --> 00:37:47
This is -- this number is e to
the i pi over three and e to the

620
00:37:47 --> 00:37:49
minus i pi over three.

621
00:37:49 --> 00:37:53
We'll be doing more complex
numbers

622
00:37:53 --> 00:37:57
briefly but a little more in
the next two days.

623
00:37:57 --> 00:37:59
next two lectures.

624
00:37:59 --> 00:38:04
Anyway, the -- so what's the
deal with stability,

625
00:38:04 --> 00:38:06
what do the Dn-s do?

626
00:38:06 --> 00:38:12
Well, look, if -- if I take the
sixth power I'm around at one,

627
00:38:12 --> 00:38:15
the problem actually told me
this.

628
00:38:15 --> 00:38:19.51
The sixth power of those
eigenvalues brings me around to

629
00:38:19.51 --> 00:38:19
one.

630
00:38:19 --> 00:38:23
What does that tell you about
the matrix, by the way?

631
00:38:23 --> 00:38:26
Suppose you know -- this was a
great quiz question,

632
00:38:26 --> 00:38:29.61
so I should never have just
said it, but popped out.

633
00:38:29.61 --> 00:38:32.57
Suppose lambda one to the sixth
and

634
00:38:32.57 --> 00:38:36
lambda two to the sixth are --
are one, which they are.

635
00:38:36 --> 00:38:39
What does that tell me about a
m- a matrix?

636
00:38:39 --> 00:38:41
About my matrix A here.

637
00:38:41 --> 00:38:46
Well, what -- what matrix is
connected with lambda one to the

638
00:38:46 --> 00:38:48
sixth and lambda two to the
sixth?

639
00:38:48 --> 00:38:52
It's got to be the matrix
A to the sixth.

640
00:38:52 --> 00:38:56.37
So what is A to the sixth for
that matrix?

641
00:38:56.37 --> 00:38:59
It's got eigenvalues one and
one.

642
00:38:59 --> 00:39:03
Because when I take the sixth
power, actually,

643
00:39:03 --> 00:39:08
ye, if I take the sixth power
b- all the sixth power of that

644
00:39:08 --> 00:39:11
is one and the sixth power of
that

645
00:39:11 --> 00:39:15
is one, the sixth power of this
is e to the two pi i,

646
00:39:15 --> 00:39:20
that's one, the sixth power of
this is e to the minus two pi i,

647
00:39:20 --> 00:39:21.23
that's one.

648
00:39:21.23 --> 00:39:24
So the sixth powers,
the -- the sixth power of that

649
00:39:24 --> 00:39:28
matrix has eigenvalues one and
one, so what is it?

650
00:39:28 --> 00:39:31
It's the identity,
right.

651
00:39:31 --> 00:39:35.59
So if I operate this -- if I
run this thing six times,

652
00:39:35.59 --> 00:39:37.31
I'm back where I was.

653
00:39:37.31 --> 00:39:41
The sixth power of that matrix
is the identity.

654
00:39:41 --> 00:39:41
Good.

655
00:39:41 --> 00:39:41
OK.

656
00:39:41 --> 00:39:45
So it'll loop around,
it's -- it doesn't go to zero,

657
00:39:45 --> 00:39:51
it doesn't blow up,
it just periodically goes

658
00:39:51 --> 00:39:53
around with period six.

659
00:39:53 --> 00:39:57
OK.
let's just see if there's a --

660
00:39:57 --> 00:39:58
all right.

661
00:39:58 --> 00:40:00
I'll -- let's see.

662
00:40:00 --> 00:40:06
Could I also look at a -- at a
final exam from nineteen

663
00:40:06 --> 00:40:07
ninety-two.

664
00:40:07 --> 00:40:12
I think that's yeah,
let me do that on this last

665
00:40:12 --> 00:40:14
board.

666
00:40:14 --> 00:40:19
It starts -- a lot of the
questions in this exam are about

667
00:40:19 --> 00:40:21
a family of matrices.

668
00:40:21 --> 00:40:26
Let me give you the fourth,
the fourth guy in the family is

669
00:40:26 --> 00:40:30
-- has a one,
so it's zeroes on the diagonal,

670
00:40:30 --> 00:40:34
but these are going one,
two, three and so on.

671
00:40:34 --> 00:40:38
One, two, three,
and so on.

672
00:40:38 --> 00:40:42
But, for the four-by-four case
I'm stopping at four.

673
00:40:42 --> 00:40:44
You see the pattern?

674
00:40:44 --> 00:40:50
It's a family of matrices which
is growing, and actually the

675
00:40:50 --> 00:40:54
numbers -- it's symmetric,
right, it's equal to A4

676
00:40:54 --> 00:40:55
transpose.

677
00:40:55 --> 00:41:00
And we can ask all sorts of
questions about its

678
00:41:00 --> 00:41:08
null space, its range,
r- its column space find the

679
00:41:08 --> 00:41:14
projection matrix onto the
column space of A3,

680
00:41:14 --> 00:41:18
for example,
is in here.

681
00:41:18 --> 00:41:24
So -- so one -- so A3 is zero,
one, zero, one,

682
00:41:24 --> 00:41:28
zero, two, zero,
two, zero.

683
00:41:28 --> 00:41:35
OK, find the projection matrix
onto

684
00:41:35 --> 00:41:36
the column space.

685
00:41:36 --> 00:41:41
By the way, is that matrix
singular or invertible?

686
00:41:41 --> 00:41:42
Singular.

687
00:41:42 --> 00:41:45
Why do we know it's singular?

688
00:41:45 --> 00:41:49
I see that column three is a
multiple of column one.

689
00:41:49 --> 00:41:52
Or we could take its
determinant.

690
00:41:52 --> 00:41:55
So it's certainly singular.

691
00:41:55 --> 00:41:59.37
The projection
will be matrix will be

692
00:41:59.37 --> 00:42:03
three-by-three but it will
project onto the column space,

693
00:42:03 --> 00:42:06
it'll project onto this plane.

694
00:42:06 --> 00:42:10
The column space of A3,
and I guess I would find it

695
00:42:10 --> 00:42:13
from the formula AA -- AA
transpose A inverse,

696
00:42:13 --> 00:42:18
I would have to -- I would -- I
guess I would do all

697
00:42:18 --> 00:42:19
this.

698
00:42:19 --> 00:42:23
There may be a better way,
perhaps I could think there

699
00:42:23 --> 00:42:27.92
might be a slightly quicker way,
but that would come out pretty

700
00:42:27.92 --> 00:42:28
fast.

701
00:42:28 --> 00:42:28.53
OK.

702
00:42:28.53 --> 00:42:31
So that's be the projection
matrix.

703
00:42:31 --> 00:42:32
Next question.

704
00:42:32 --> 00:42:36
Find the eigenvalues and
eigenvectors of that matrix.

705
00:42:36 --> 00:42:36.99
OK.

706
00:42:36.99 --> 00:42:39
There's a three-by-three
matrix, oh, yeah,

707
00:42:39 --> 00:42:42
so what are its eigenvalues and
eigenvectors,

708
00:42:42 --> 00:42:45
we haven't done any
three-by-threes.

709
00:42:45 --> 00:42:46
Let's do one.

710
00:42:46 --> 00:42:49
I want to find,
so how do I find eigenvalues?

711
00:42:49 --> 00:42:53
I take the determinant of A3
minus lambda I.

712
00:42:53 --> 00:42:58
So this is you just have to --
so I'm subtracting lambda from

713
00:42:58 --> 00:43:00
the diagonal,
and I have a one,

714
00:43:00 --> 00:43:03
one, zero, zero,
two, two there,

715
00:43:03 --> 00:43:06.82
and I just have to find that
determinant.

716
00:43:06.82 --> 00:43:11
OK, since it's three-by-three
I'll just go for it.

717
00:43:11 --> 00:43:16
This way gives me minus lambda
cubed and a zero and zero.

718
00:43:16 --> 00:43:20
Then in this direction which
has the minus sign,

719
00:43:20 --> 00:43:22
that's a zero,
four lambdas,

720
00:43:22 --> 00:43:27
I mean minus four lambdas,
and minus another lambda,

721
00:43:27 --> 00:43:32
so that's minus five lambdas,
but that direction goes with a

722
00:43:32 --> 00:43:36.66
minus sign, so I think it's plus
five lambda.

723
00:43:36.66 --> 00:43:41
That looks like the determinant
of A3 minus lambda I,

724
00:43:41 --> 00:43:42
so I set it to zero.

725
00:43:42 --> 00:43:45.09
So what are the eigenvalues?

726
00:43:45.09 --> 00:43:49
Well, lambda equals zero --
lambda factors out of this,

727
00:43:49 --> 00:43:53.82
times minus lambda
squared plus four,

728
00:43:53.82 --> 00:43:58
so the eigenvalues are five,
thanks, thanks,

729
00:43:58 --> 00:44:03
so the eigenvalues are zero,
square root of five,

730
00:44:03 --> 00:44:06
and minus square root of five.

731
00:44:06 --> 00:44:11
And I would never write down
those three eigenvalues without

732
00:44:11 --> 00:44:15
checking the trace to tell the
truth.

733
00:44:15 --> 00:44:20
Because -- because we did
a bunch of calculations here

734
00:44:20 --> 00:44:24
but then I can quickly add up
the eigenvalues to get zero,

735
00:44:24 --> 00:44:27
add up the trace to get zero,
and feel that I'm -- well,

736
00:44:27 --> 00:44:30
I guess that wouldn't have
caught my error if I'd made it

737
00:44:30 --> 00:44:34
-- if -- if that had been
a four I wouldn't have

738
00:44:34 --> 00:44:39
noticed,the determinant isn't
anything greatly useful here,

739
00:44:39 --> 00:44:42
right, because the determinant
is just zero.

740
00:44:42 --> 00:44:47
And so I never would know
whether that five was right or

741
00:44:47 --> 00:44:50
wrong, but thanks for making it
right.

742
00:44:50 --> 00:44:50
OK.

743
00:44:50 --> 00:44:52.17
Ha.

744
00:44:52.17 --> 00:44:56
Question two c,
whoever wrote this,

745
00:44:56 --> 00:45:00
probably me,
said this is not difficult.

746
00:45:00 --> 00:45:06
I don't know why I put that in.
just -- it asks for the

747
00:45:06 --> 00:45:12
projection matrix onto the
column space of A4.

748
00:45:12 --> 00:45:18
How could I have thought that
wasn't difficult?

749
00:45:18 --> 00:45:24
It looks extremely difficult.
what's the projection matrix

750
00:45:24 --> 00:45:28
onto the column space of A4?

751
00:45:28 --> 00:45:34
I don't know whether that --
this is not difficult is just

752
00:45:34 --> 00:45:37
like helpful or -- or insulting.

753
00:45:37 --> 00:45:40
Uh, what do you think?

754
00:45:40 --> 00:45:44
The -- what's the column space
of A4 here?

755
00:45:44 --> 00:45:49
Well, what's our first question
is

756
00:45:49 --> 00:45:52
is the matrix singular or
invertible?

757
00:45:52 --> 00:45:57
If the answer is invertible,
then what's the column space?

758
00:45:57 --> 00:46:02
If -- if this matrix A4 is
invertible, so that's my guess,

759
00:46:02 --> 00:46:07.2
if this problem's easy it has
to be because this matrix is

760
00:46:07.2 --> 00:46:09
probably invertible.

761
00:46:09 --> 00:46:14
Then its column space is R^4,
good, the column space is the

762
00:46:14 --> 00:46:18
whole space, and the answer to
this easy question is the

763
00:46:18 --> 00:46:22
projection matrix is the
identity, it's the four-by-four

764
00:46:22 --> 00:46:23
identity matrix.

765
00:46:23 --> 00:46:25
If this matrix is invertible.

766
00:46:25 --> 00:46:28
Shall we check invertibility?

767
00:46:28 --> 00:46:30
How would you find its
determinant?

768
00:46:30 --> 00:46:34
Can we just like take the
determinant of that matrix?

769
00:46:34 --> 00:46:38
I could ask you how -- so there
-- there are twenty-four terms,

770
00:46:38 --> 00:46:40
do we want to write all
twenty-four terms down?

771
00:46:40 --> 00:46:42
not in the remaining ten
seconds.

772
00:46:42 --> 00:46:45
Better to use cofactors.

773
00:46:45 --> 00:46:49
So I go along row one,
I see one -- the only nonzero

774
00:46:49 --> 00:46:53.41
is this guy, so I should take
that one times the cofactor.

775
00:46:53.41 --> 00:46:56
Now so I'm down to this
determinant.

776
00:46:56 --> 00:46:56
OK.

777
00:46:56 --> 00:46:59
So now I'm -- look at this
first column,

778
00:46:59 --> 00:47:02
I see one times this,
there's the

779
00:47:02 --> 00:47:06
cofactor of the one,
so I'm using up row one -- row

780
00:47:06 --> 00:47:09
one and column one of this
three-by-three matrix,

781
00:47:09 --> 00:47:12
I'm down to this cofactor,
and by the way,

782
00:47:12 --> 00:47:14
those were both plus signs,
right?

783
00:47:14 --> 00:47:16
No, they weren't.

784
00:47:16 --> 00:47:17
That was a minus sign.

785
00:47:17 --> 00:47:21
That was a --
that was a minus,

786
00:47:21 --> 00:47:24.98
and then that was a plus,
and then this,

787
00:47:24.98 --> 00:47:27
so what's the determinant?

788
00:47:27 --> 00:47:28
Nine.

789
00:47:28 --> 00:47:28
Nine.

790
00:47:28 --> 00:47:30
Determinant is nine.

791
00:47:30 --> 00:47:32
Determinant of A4 is nine.

792
00:47:32 --> 00:47:33
OK.

793
00:47:33 --> 00:47:39
Where A3, so my guess is I'll
put that on the final this year,

794
00:47:39 --> 00:47:45
the -- probably the odd-
numbered ones are singular and

795
00:47:45 --> 00:47:49.14
the even-numbered ones are
invertible.

796
00:47:49.14 --> 00:47:54
And I don't know what the
determinants are but I'm betting

797
00:47:54 --> 00:47:57
that they have some nice
formula.

798
00:47:57 --> 00:47:57
OK.

799
00:47:57 --> 00:48:03
So, recitations this week will
also be quiz review and then the

800
00:48:03 --> 00:48:06
quiz is Wednesday at one
o'clock.

801
00:48:06 --> 00:48:09
Thanks.