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PROFESSOR: Hi, everyone.
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Today, we're going to talk
about linear transformations.
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So, we've seen linear
transformations incognito
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all the time until now.
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We've played around
with matrices.
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Matrices multiplying
vectors, say,
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in R^n and producing
vectors in R^m.
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So really, the language
of linear transformations
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only provides a
nicer framework when
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we want to analyze
linear operations on more
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abstract vector spaces, like
the one we have in this problem
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here.
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We're going to work with the
space of two by two matrices.
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And we're going to analyze the
operation, have the matrix A,
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and we produce its transpose.
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OK.
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So please take a few minutes
to try the problem on your own
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and come back.
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Hi, again.
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OK.
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So the first question
we need to ask ourselves
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is, indeed, why is
T a linear operator?
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So what are the
abstract properties
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that a linear
operator satisfies?
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Well, what happens when T acts
on the sum of two matrices, A
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and B?
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So it produces the matrix
the transpose of A plus B.
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But we know that this is
A transpose, B transpose.
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And so, this is
exactly T(A) plus T(B).
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So the transformation
that we're analyzing
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takes the sum of two
matrices into the sum
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of their transformations.
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OK.
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Similarly, it takes a
multiple of a transformation
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into the multiple of
the transformations.
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So it takes the matrix c times
A to c times A transpose, which
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is c T of A.
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OK.
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So it is a linear operator.
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Now, can we figure out
what its inverse is?
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Well, what does
the transpose do?
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It takes a column and
flips it into a row.
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So what happens if we apply
the operation once again?
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Well, it's going to
take the row and turn it
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back down to the column.
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So applying the
transformation twice,
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we come back to the
original situation.
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So therefore, T squared
is the identity.
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And from this, we
infer that the inverse
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is the transformation itself.
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Now, this was part one.
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Part two, we'll
compute the matrix
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of the linear transformation
in the following two bases.
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So the first basis
is, in fact-- it
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is the standard basis for the
space of two by two matrices.
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And the way we compute
the matrix, we first
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compute what T does to
each of the basis elements.
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So T of v_1.
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Let's go back.
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So here.
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So T takes the transpose
of this matrix.
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And we see that the transpose
of [1, 0; 0, 0] is [1, 0; 0, 0].
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So it's a symmetric matrix.
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So T of v_1 is v_1.
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What about T of v_2?
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Come back here.
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So this 1 comes here.
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0 comes here.
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And so we actually get v_3.
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So T of v_2 is v_3.
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Similarly, T of v_3 is v_2.
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And finally, T of v_4.
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Well, v_4 is a symmetric
matrix as well.
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So the transpose
doesn't change it.
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OK.
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Now, we encode this into a
matrix in the following way.
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Essentially, the first column
will tell us how T of v_1
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is expressed as a linear
combination of the basis
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elements.
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Well, in this case,
it's just v_1.
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So it's going to be 1 times
v_1 plus 0*v_2 plus 0*v_3 plus
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0*v_4.
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T of v_2 is v_3.
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So we have 0, 0, 1, 0.
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T of v_3 is 0*v_1,
1*v_2, 0*v_3, 0*v_4.
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And T of v4 is 0*v_1,
0*v_2, 0*v_3, plus 1*v_4.
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OK.
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So we've written down the matrix
of the linear transformation T
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in the standard basis.
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And you can check that this
is exactly what we want.
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The representation of some
matrix, say, [1, 2; 3, 4]
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in this standard basis is,
it's the vector [1, 2, 3, 4].
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T takes this to its
transpose, [1, 3; 2, 4].
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So this in the basis is
represented as [1, 3, 2, 4].
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Right?
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And it's not hard
to see that when
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M_T multiplies this vector,
we get exactly this vector.
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So we'll pause for a bit,
so that I erase the board.
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And we're going to return
with the representation of T
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in the basis w_1,
w_2, w_3, and w_4.
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OK.
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So let's now
compute the matrix T
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in the basis w_1,
w_2, w_3, and w_4.
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We played the same game.
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We look at how T acts on
each of the basis vectors.
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So T of w_1-- well, w_1
is a symmetric matrix.
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So T of w_1 is w_1.
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Similarly, with w_2 and w_3.
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They're all symmetric.
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What about w_4?
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Well, we see that the
1 comes down here,
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the negative one comes
up here, and in the end,
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we just get the negative of w_4.
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So, let me just write this out.
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We had T of w_1 equal to
w_1, T of w_2 equal to w_2,
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T of w_3 equal to w_3, and T
of w_4, was negative of w_4.
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So therefore, the matrix of
the linear transformation
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T, in this basis-- I'm going
to call the matrix M prime T--
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has a fairly simple expression.
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The only non-zero entries
are on a diagonal.
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And they're precisely
1, 1, 1, and negative 1.
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And finally, let's
tackle the eigenvalues
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slash eigenvectors issue.
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Well, you've seen what an
eigenvector for a matrix is.
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And the idea for an
eigenvalue, eigenvector
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for a linear transformation
is virtually the same.
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And we are looking for the
vectors v and scalars lambda
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such that T of v is lambda*v.
But if you guys look back
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to what we just did with
w_1, w_2, w_3, and w_4,
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you'll see precisely
that w_1, w_2,
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and w_3 are eigenvectors
for T with eigenvalue 1.
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And w_4 is an eigenvector for
T with eigenvalue negative 1.
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So yeah, we essentially have
solved the problem knowing
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a very, very nice
basis in which we
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computed the linear
transformation T.
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So I'll leave it at that.