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# Example solving for the eigenvalues of a 2x2 matrix

Video transcript

In the last video we were able
to show that any lambda that satisfies this equation for some
non-zero vectors, V, then the determinant of lambda times
the identity matrix minus A, must be equal to 0. Or if we could rewrite this as
saying lambda is an eigenvalue of A if and only if-- I'll
write it as if-- the determinant of lambda times the
identity matrix minus A is equal to 0. Now, let's see if we can
actually use this in any kind of concrete way to figure
out eigenvalues. So let's do a simple 2
by 2, let's do an R2. Let's say that A is equal to
the matrix 1, 2, and 4, 3. And I want to find the
eigenvalues of A. So if lambda is an eigenvalue
of A, then this right here tells us that the determinant
of lambda times the identity matrix, so it's going to be
the identity matrix in R2. So lambda times 1, 0, 0, 1,
minus A, 1, 2, 4, 3, is going to be equal to 0. Well what does this equal to? This right here is
the determinant. Lambda times this is just lambda
times all of these terms. So it's lambda times 1
is lambda, lambda times 0 is 0, lambda times 0 is 0, lambda
times 1 is lambda. And from that we'll
subtract A. So you get 1, 2, 4, 3, and
this has got to equal 0. And then this matrix, or this
difference of matrices, this is just to keep the
determinant. This is the determinant of. This first term's going
to be lambda minus 1. The second term is 0 minus
2, so it's just minus 2. The third term is 0 minus
4, so it's just minus 4. And then the fourth term
is lambda minus 3, just like that. So kind of a shortcut to
see what happened. The terms along the diagonal,
well everything became a negative, right? We negated everything. And then the terms around
the diagonal, we've got a lambda out front. That was essentially the
byproduct of this expression right there. So what's the determinant
of this 2 by 2 matrix? Well the determinant of this is
just this times that, minus this times that. So it's lambda minus 1, times
lambda minus 3, minus these two guys multiplied
by each other. So minus 2 times minus 4
is plus eight, minus 8. This is the determinant of this
matrix right here or this matrix right here, which
simplified to that matrix. And this has got to
be equal to 0. And the whole reason why that's
got to be equal to 0 is because we saw earlier,
this matrix has a non-trivial null space. And because it has a non-trivial
null space, it can't be invertible and
its determinant has to be equal to 0. So now we have an interesting
polynomial equation right here. We can multiply it out. We get what? Let's multiply it out. We get lambda squared, right,
minus 3 lambda, minus lambda, plus 3, minus 8,
is equal to 0. Or lambda squared, minus
4 lambda, minus 5, is equal to 0. And just in case you want to
know some terminology, this expression right here is known
as the characteristic polynomial. Just a little terminology,
polynomial. But if we want to find the
eigenvalues for A, we just have to solve this right here. This is just a basic
quadratic problem. And this is actually
factorable. Let's see, two numbers and you
take the product is minus 5, when you add them
you get minus 4. It's minus 5 and plus 1, so you
get lambda minus 5, times lambda plus 1, is equal
to 0, right? Minus 5 times 1 is minus 5, and
then minus 5 lambda plus 1 lambda is equal to
minus 4 lambda. So the two solutions of our
characteristic equation being set to 0, our characteristic
polynomial, are lambda is equal to 5 or lambda is
equal to minus 1. So just like that, using the
information that we proved to ourselves in the last video,
we're able to figure out that the two eigenvalues of A are
lambda equals 5 and lambda equals negative 1. Now that only just solves part
of the problem, right? We know we're looking
for eigenvalues and eigenvectors, right? We know that this equation can
be satisfied with the lambdas equaling 5 or minus 1. So we know the eigenvalues, but
we've yet to determine the actual eigenvectors. So that's what we're going
to do in the next video.