Hey guys, in this video I'm going to do another proof problem that you would see in a linear
algebra class. This particular problem says let A be an--I don't know how you pronounce
that--impotent n by n matrix. Prove that 0 and 1 are the only possible eigenvalues of A. When
I first read this problem, I didn't immediately think that was an obvious conclusion to draw but
it is a pretty straightforward proof. I think this will be a short video. Let's go ahead and write
out the things that we're given as assumptions. We can say by assumption A is an idempotent n by
n matrix. We know that and the definition of that is basically that if you multiply A by A you still
get A. So we can say therefore A squared equals A. Now suppose lambda is an eigenvalue of A,
which we know the matrix A is going to have an eigenvalue. We have to basically--it's asking
us already to prove that the eigenvalues are 0 or 1. Then there exists a non zero vector V such that
A lambda equals--sorry, AV equals lambda V. This is one of the most basic definitions of matrices.
When you're--it's kind of like the first thing you learn when you're learning about eigenvalues
and eigenvectors that that's the definition of what it means to have an eigenvector or to have
an eigenvalue. So we can try to use this equation because we know we're going to try to solve for
lambda. Another equation I thought of was the finding the determinant and that also involves
finding lambda but this is a more straightforward way of solving this proof. So okay, we have that.
Well we know that A squared equals A, so one thing we can do is multiply both sides by A since it's
an n by n matrix you can do that and you're going to get A squared on one side. So let's go ahead
and multiply A on both sides just to see what we get. That equals A squared V equal to a lambda V
and then we're just going to play around with this to get just eigenvalues. So just lambda.
We know that AV equals lambda V, so we have AV here. We can pull the eigenvalue out because
that's just a scalar. We can say that's equal to lambda AV. Put parentheses around that then
we know that A squared V is equal to lambda and then this is equal to lambda V. We have
lambda times lambda V. We successfully got just lambda squared V on the right hand side. We
still have A squared V on the left hand side. A squared V is equivalent to AV we know that,
which is equal to lambda V, which is equal to lambda squared V. And we can say that because
of the property that A is idempotent. You can write that on the side if that's necessary. Then
if we use this as the left hand side and this is the right hand side, we have an equation that we
can solve. So we'll have lambda V equal to lambda squared V. Then we can move that all to one side.
We have 0 equal to lambda squared V minus lambda V and if we pull out the eigenvector, we'll have
lambda squared minus lambda equals to 0. Since by definition we said at the top that the eigenvalue
cannot be 0 by definition--Since V is nonzero by definition--therefore we have 0 is equal to lambda
squared minus lambda. If we pull out a lambda, we get lambda times lambda minus 1 equal to 0 and
finally with this last line we're able to conclude that either the eigenvalue is 0 or the eigenvalue
is 1 and that's enough for our proof. It wasn't a very difficult proof, you basically just had to
remember the simple definition of what it means to have an eigenvalue and an eigenvector for a
matrix and you had to know what impotence meant but besides that it was pretty straightforward.