WEBVTT

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In the last video we were able
to show that any lambda that

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satisfies this equation for some
non-zero vectors, V, then

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the determinant of lambda times
the identity matrix

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minus A, must be equal to 0.

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Or if we could rewrite this as
saying lambda is an eigenvalue

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of A if and only if-- I'll
write it as if-- the

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determinant of lambda times the
identity matrix minus A is

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equal to 0.

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Now, let's see if we can
actually use this in any kind

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of concrete way to figure
out eigenvalues.

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So let's do a simple 2
by 2, let's do an R2.

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Let's say that A is equal to
the matrix 1, 2, and 4, 3.

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And I want to find the
eigenvalues of A.

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So if lambda is an eigenvalue
of A, then this right here

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tells us that the determinant
of lambda times the identity

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matrix, so it's going to be
the identity matrix in R2.

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So lambda times 1, 0, 0, 1,
minus A, 1, 2, 4, 3, is going

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to be equal to 0.

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Well what does this equal to?

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This right here is
the determinant.

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Lambda times this is just lambda
times all of these

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terms. So it's lambda times 1
is lambda, lambda times 0 is

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0, lambda times 0 is 0, lambda
times 1 is lambda.

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And from that we'll
subtract A.

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So you get 1, 2, 4, 3, and
this has got to equal 0.

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And then this matrix, or this
difference of matrices, this

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is just to keep the
determinant.

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This is the determinant of.

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This first term's going
to be lambda minus 1.

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The second term is 0 minus
2, so it's just minus 2.

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The third term is 0 minus
4, so it's just minus 4.

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And then the fourth term
is lambda minus

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3, just like that.

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So kind of a shortcut to
see what happened.

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The terms along the diagonal,
well everything became a

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negative, right?

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We negated everything.

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And then the terms around
the diagonal, we've got

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a lambda out front.

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That was essentially the
byproduct of this expression

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right there.

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So what's the determinant
of this 2 by 2 matrix?

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Well the determinant of this is
just this times that, minus

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this times that.

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So it's lambda minus 1, times
lambda minus 3, minus these

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two guys multiplied
by each other.

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So minus 2 times minus 4
is plus eight, minus 8.

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This is the determinant of this
matrix right here or this

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matrix right here, which
simplified to that matrix.

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And this has got to
be equal to 0.

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And the whole reason why that's
got to be equal to 0 is

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because we saw earlier,
this matrix has a

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non-trivial null space.

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And because it has a non-trivial
null space, it

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can't be invertible and
its determinant has

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to be equal to 0.

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So now we have an interesting
polynomial

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equation right here.

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We can multiply it out.

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We get what?

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Let's multiply it out.

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We get lambda squared, right,
minus 3 lambda, minus lambda,

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plus 3, minus 8,
is equal to 0.

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Or lambda squared, minus
4 lambda, minus

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5, is equal to 0.

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And just in case you want to
know some terminology, this

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expression right here is known
as the characteristic

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polynomial.

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Just a little terminology,
polynomial.

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But if we want to find the
eigenvalues for A, we just

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have to solve this right here.

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This is just a basic
quadratic problem.

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And this is actually
factorable.

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Let's see, two numbers and you
take the product is minus 5,

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when you add them
you get minus 4.

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It's minus 5 and plus 1, so you
get lambda minus 5, times

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lambda plus 1, is equal
to 0, right?

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Minus 5 times 1 is minus 5, and
then minus 5 lambda plus 1

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lambda is equal to
minus 4 lambda.

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So the two solutions of our
characteristic equation being

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set to 0, our characteristic
polynomial, are lambda is

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equal to 5 or lambda is
equal to minus 1.

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So just like that, using the
information that we proved to

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ourselves in the last video,
we're able to figure out that

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the two eigenvalues of A are
lambda equals 5 and lambda

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equals negative 1.

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Now that only just solves part
of the problem, right?

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We know we're looking
for eigenvalues and

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eigenvectors, right?

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We know that this equation can
be satisfied with the lambdas

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equaling 5 or minus 1.

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So we know the eigenvalues, but
we've yet to determine the

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actual eigenvectors.

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So that's what we're going
to do in the next video.

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