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GILBERT STRANG: So today begins
eigenvalues and eigenvectors.

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And the reason we
want those, need

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those is to solve systems
of linear equations.

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Systems meaning more than
one equation, n equations.

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n equal 2 in the examples here.

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So eigenvalue is a number,
eigenvector is a vector.

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They're both hiding
in the matrix.

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Once we find them,
we can use them.

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Let me show you the reason
eigenvalues were created,

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invented, discovered was solving
differential equations, which

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is our purpose.

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So why is now a vector-- so
this is a system of equations.

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I'll do an example in a minute.

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A is a matrix.

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So we have n equations,
n components of y.

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And A is an n by n
matrix, n rows, n columns.

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Good.

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And now I can tell you
right away where eigenvalues

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and eigenvectors pay off.

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They come into the solution.

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We look for solutions
of that kind.

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When we had one equation,
we looked for solutions

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just e to the st, and
we found that number s.

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Now we have e to the
lambda t-- we changed s

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to lambda, no problem--
but multiplied by a vector

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because our unknown is a vector.

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This is a vector, but that
does not depend on time.

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That's the beauty of it.

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All the time dependence is in
the exponential, as always.

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And x is just multiples of that
exponential, as you'll see.

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So I look for
solutions like that.

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I plug that into the
differential equation

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and what happens?

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So here's my equation.

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I'm plugging in what e to
the lambda tx, that's y.

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That's A times y there.

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Now, the derivative of
y, the time derivative,

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brings down a lambda.

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To get the derivative
I include the lambda.

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So do you see that
substituting into the equation

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with this nice notation is
just this has to be true.

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My equation changed
to that form.

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OK Now I cancel either
the lambda t, just the way

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I was always
canceling e to the st.

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So I cancel e to the lambda
t because it's never zero.

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And I have the big
equation, Ax, the matrix

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times my eigenvector,
is equal to lambda

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x-- the number, the eigenvalue,
times the eigenvector.

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Not linear, notice.

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Two unknowns here
that are multiplied.

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A number, lambda,
times a vector, x.

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So what am I looking for?

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I'm looking for vectors x,
the eigenvectors, so that

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multiplying by A-- multiplying A
times x gives a number times x.

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It's in the same direction as
x just the length is changed.

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Well, if lambda was 1,
I would have Ax equal x.

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That's allowed.

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If lambda is 0, I
would have Ax equals 0.

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That's all right.

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I don't want x to be 0.

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That's useless.

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That's no help to know
that 0 is a solution.

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So x should be not 0.

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Lambda can be any number.

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It can be real, it
could be complex number,

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as you will see.

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Even if the matrix is real,
lambda could be complex.

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Anyway, Ax equal lambda x.

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That's the big equation.

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It got a box around it.

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So now I'm ready
to do an example.

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And in this example,
first of all,

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I'm going to spot the
eigenvalues and eigenvectors

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without a system, just go
for it in the 2 by 2 case.

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So I'll give a 2 by 2 matrix
A. We'll find the lambdas

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and the x's, and then we'll
have the solution to the system

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of differential equations.

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Good.

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There's the system.

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There's the first equation for
y1-- prime meaning derivative,

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d by dt, time derivative-- is
linear, a constant coefficient.

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Second one, linear, constant
coefficient, 3 and 3.

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Those numbers, 5, 1, 3,
3, go into the matrix.

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Then that problem is exactly y
prime, the vector, derivative

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of the vector, equal A times y.

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That's my problem.

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Now eigenvalues and
eigenvectors will solve it.

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So I just look at that matrix.

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Matrix question.

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What are the eigenvalues,
what are the eigenvectors

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of that matrix?

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And remember, I want
Ax equals lambda x.

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I've spotted the
first eigenvector.

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1, 1.

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We could just
check does it work.

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If I multiply A by
that eigenvector,

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1, 1, do you see what
happens when I multiply by 1?

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That gives me a 6.

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That gives me a 6.

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So A times that vector is 6, 6.

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And that is 6 times 1, 1.

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So there you go.

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Found the first eigenvalue.

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If I multiply A by
x, I get 6 by x.

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I get the vector 6, 6.

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Now, the second one.

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Again, I've worked in advance,
produced this eigenvector,

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and I think it's 1 minus 3.

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So let's multiply by A.
Try the second eigenvector.

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I should call this first
one maybe x1 and lambda 1.

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And I should call this
one x2 and lambda 2.

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And we can find out what
lambda 2 is, once I find

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the eigenvectors of course.

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I just do A times x to recognize
the lambda, the eigenvalue.

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So 5, 1 times this
is 5 minus 3 is a 2.

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It's a 2.

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So here I got a 2.

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And from 3, 3 it's 3
minus 9 is minus 6.

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That's what I got for Ax.

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There was the x.

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When I did the multiplication,
Ax came out to be 2 minus 6.

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Good.

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That output is two
times the input.

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The eigenvalue is 2.

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Right?

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I'm looking for inputs,
the eigenvector,

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so that the output is a
number times that eigenvector,

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and that number is
lambda, the eigenvalue.

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So I've now found the two.

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And I expect two
for a 2 by 2 matrix.

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You will soon see why I
expect two eigenvalues,

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and each eigenvalue should
have an eigenvector.

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So here they are
for this matrix.

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So I've got the answers now.

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y of t, which stands
for y1 and y2 of t.

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Those are-- it's e
to the lambda tx.

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Remember, that's the picture
that we're looking for.

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So the first one is e to the
6t times x, which is 1, 1.

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If I put that into the equation,
it will solve the equation.

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Also, I have another one.

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e to the lambda 2 was 2t.

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e to the lambda t times
its eigenvector, 1 minus 3.

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That's a solution also.

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One solution, another solution.

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And what do I do with
linear equations?

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I take combinations.

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Any number c1 of that,
plus any number c2 of that

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is still a solution.

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That's superposition, adding
solutions to linear equations.

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These are null equations.

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There's no force term
in these equations.

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I'm not dealing
with a force term.

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I'm looking for the null
solutions, the solutions

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of the equations themselves.

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And there I have two solutions,
two coefficients to choose.

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How do I choose them?

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Of course, I match the initial
condition, so at t equals 0.

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At t equals 0.

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At t equals 0, I
would have y of 0.

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That's my given initial
condition, my y1 and y2.

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So I'm setting t equals 0,
so that's one of course.

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When t is 0, that's one.

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So I just have c1 times 1, 1.

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And c2-- that's one again at
t equals o-- times 1 minus 3.

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That's what
determines c1 and c2.

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c1 and c2 come from
the initial conditions

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just the way they always did.

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So I'm solving two first order
linear constant coefficient

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equations, homogeneous,
meaning no force term.

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So I get a null solution
with constants to choose

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and, as always, those
constants come from matching

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the initial conditions.

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So the initial condition
here is a vector.

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So if, for example, y
of 0 was 2 minus 2, then

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I would want one of
those and one of those.

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OK.

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I've used eigenvalues
and eigenvectors

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to solve a linear system, their
first and primary purpose.

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OK.

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But how do I find those
eigenvalues and eigenvectors?

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What about other properties?

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What's going on with
eigenvalues and eigenvectors?

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May I begin on this just
a couple more minutes

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about eigenvalues
and eigenvectors?

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Basic facts and then I'll come
next video of how to find them.

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OK, basic facts.

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Basic facts.

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So start from Ax
equals lambda x.

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Let's suppose we found those.

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Could you tell me the
eigenvalues and eigenvectors

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of A squared?

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I would like to know what the
eigenvalues and eigenvectors

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of A squared are.

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Are they connected with these?

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So suppose I know the x and
I know the lambda for A. What

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about for A squared?

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Well, the good thing is
that the eigenvectors

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are the same for A squared.

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So let me show you.

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I say that same x, so this
is the same x, same vector,

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same eigenvector.

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The eigenvalue would be
different, of course,

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for A squared, but the
eigenvector is the same.

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And let's see what
happens for A squared.

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So that's A times Ax, right?

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One A, another Ax.

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But Ax is lambda x.

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Are you good with that?

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That's just A times Ax.

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So that's OK.

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Now lambda is a number.

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I like to bring it out
front where I can see it.

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So I didn't do anything there.

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This number lambda was
multiplying everything

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so I put it in front.

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Now Ax.

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I have, again, the Ax.

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That's, again, the
lambda x because I'm

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looking at the same x.

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Same x, so I get
the same lambda.

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So that's a lambda
x, another lambda.

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I have lambda squared x.

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That's what I wanted.

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A squared x is lambda squared x.

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Conclusion.

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The eigenvectors stay the same,
lambda goes to lambda squared.

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The eigenvalues are squared.

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So if I had my example again--
oh, let me find that matrix.

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Suppose I had that
same matrix and I

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was interested in
A squared, then

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the eigenvalues would be
36 and 4, the squares.

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I suppose I'm looking at
the n-th power of a matrix.

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You may say why look
at the n-th power?

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But there are many
examples to look

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at the n-th power of a
matrix, the thousandth power.

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So let's just write
down the conclusion.

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Same reasoning, A to
the n-th x is lambda.

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It's the same x.

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And every time I multiply by
A, I multiply by a lambda.

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So I get lambda n times.

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So there is the handy rule.

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And that really tells
us something about what

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eigenvalues are good for.

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Eigenvalues are good for
things that move in time.

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Differential equations, that
is really moving in time.

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n equal 1 is this first time,
or n equals 0 is the start.

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Take one step to n equal 1,
take another step to n equal 2.

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Keep going.

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Every time step brings a
multiplication by lambda.

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So that is a very useful rule.

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Another handy rule is what
about A plus the identity?

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Suppose I add the identity
matrix to my original matrix.

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00:16:23,870 --> 00:16:25,620
What happens to the eigenvalues?

254
00:16:25,620 --> 00:16:27,290
What happens to
the eigenvectors?

255
00:16:27,290 --> 00:16:28,740
Basic question.

256
00:16:28,740 --> 00:16:33,030
Or I could multiply a constant
times the identity, 2 times

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00:16:33,030 --> 00:16:35,820
the identity, 7
times the identity.

258
00:16:35,820 --> 00:16:39,890
And I want to know what
about its eigenvectors.

259
00:16:39,890 --> 00:16:45,260
And the answer is
same, same x's.

260
00:16:45,260 --> 00:16:47,950
Same x.

261
00:16:47,950 --> 00:16:53,010
I show that by figuring
out what I have here.

262
00:16:53,010 --> 00:16:57,630
This is Ax, which is lambda x.

263
00:16:57,630 --> 00:17:00,850
And this is c times
the identity times x.

264
00:17:00,850 --> 00:17:05,760
The identity doesn't do
anything so that's just cx.

265
00:17:05,760 --> 00:17:08,430
So what do I have now?

266
00:17:08,430 --> 00:17:16,319
I've seen that the
eigenvalue is lambda plus c.

267
00:17:16,319 --> 00:17:19,491
So there is the eigenvalues.

268
00:17:24,685 --> 00:17:30,100
I think about this as shifting
A by a multiple of the identity.

269
00:17:30,100 --> 00:17:34,830
Shifting A, adding 5
times the identity to it.

270
00:17:34,830 --> 00:17:38,780
If I add 5 times the
identity to any matrix,

271
00:17:38,780 --> 00:17:43,710
the eigenvalues of
that matrix go up by 5.

272
00:17:43,710 --> 00:17:46,800
And the eigenvectors
stay the same.

273
00:17:46,800 --> 00:17:51,650
So as long as I keep working
with that one matrix A.

274
00:17:51,650 --> 00:17:57,650
Taking powers, adding
multiples of the identity,

275
00:17:57,650 --> 00:18:02,050
later taking exponentials,
whatever I do I keep

276
00:18:02,050 --> 00:18:05,730
the same eigenvectors
and everything is easy.

277
00:18:10,910 --> 00:18:17,390
If I had two matrices, A and
B, with different eigenvectors,

278
00:18:17,390 --> 00:18:21,280
then I don't know what the
eigenvectors of A plus B

279
00:18:21,280 --> 00:18:22,330
would be.

280
00:18:22,330 --> 00:18:24,430
I don't know those.

281
00:18:24,430 --> 00:18:28,040
I can't tell the
eigenvectors of A times B

282
00:18:28,040 --> 00:18:30,610
because A has its own
little eigenvectors

283
00:18:30,610 --> 00:18:32,420
and B has its eigenvectors.

284
00:18:32,420 --> 00:18:38,310
Unless they're the same, I
can't easily combine A and B.

285
00:18:38,310 --> 00:18:47,770
But as always I'm staying with
one A and its powers and steps

286
00:18:47,770 --> 00:18:50,090
like that, no problem.

287
00:18:50,090 --> 00:18:50,720
OK.

288
00:18:50,720 --> 00:18:54,240
I'll stop there for a
first look at eigenvalues

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00:18:54,240 --> 00:18:55,990
and eigenvectors.