All right, so in this problem, we are going to look at some matrices and determine if
some given vectors are eigenvectors of the matrix or not.
So in part a, let's deal with this 2 by 2 matrix A right here.
So A is the matrix 3 negative 3 2 negative 4.
And we're given two vectors.
V1 is the vector 3 1, and V2 is the vector 2 1.
And the question that we are asked is, are the vectors of v1 and v2 eigenvectors of A?
So remember what an eigenvector is.
An eigenvector of a matrix is a special type of matrix.
It's a matrix that when acted on by the matrix, we get out the original vector just multiplied
by some number.
So in this problem, since we're given some vectors to check, really we're just checking
to see if when I act on this vector, do I get out my original vector scaled by a constant
or not?
So it really just comes down to doing some simple computations.
So let's go ahead and just compute what the matrix vector product Av1 is.
So if we do that, that's the matrix A times the vector v1.
And if you do that multiplication, you get the vector 6 comma 2.
So 6 2, which now I can factor out of 2.
2 times 3 is 6.
2 times 1 is 2.
So sitting right here, I have the vector v1.
I have a constant times it.
So since I was able to write it as 2 times v1, this by definition is an eigenvector.
So that's what an eigenvector is.
When my matrix acts on the eigenvector or the vector, what comes out is the vector just
times a number.
So v1 is indeed an eigenvector of the matrix A. Actually, we now know that it has an eigenvalue
of 2 associated with it, because the scale factor 2 right here was the number that was
multiplying the vector v1.
All right, let's do another one.
What about the vector v2?
Let's try that.
So let's compute what A times v2 is.
So it's this matrix times v2.
And if you work that out, you get 3 comma 0.
The vector of 3 0.
And this is definitely not equal to some number times v2.
And usually one of the tells is if I have a 0 here, there is no way I can multiply zero
by to get to be equal to 1.
There is no constant I can find y lambda 2 times as vector to yield 3 0.
There's no way I can write this math with that constant, so the vector v2 we conclude
is definitely not an eigenvector of the matrix A.
So that's how we check for eigenvectors.
If we have a matrix, if we have a vector checking for them is pretty easy, can I factor it like
this?
If I can is.
Can I not factor like this?
If you can't factor it, then it is not an eigenvector.
All right, let's do another examples, do a 3 by 3.
So here is the matrix A, and we are given some vectors v1 and v2.
And again, the same question, are the vectors v1 and v2 eigenvectors of A?
So again, let's do our computations.
Let's compute A times of v1 that is equal to this matrix right here.
And there is no number lambda 1 such that we can write this vector as a constant times
our original starting vector.
And again, the fact that we have a 0 here and something non-zero here makes that pretty
obvious.
So that means v1 is not an eigenvector of the matrix, so that is not an eigenvector.
Let's compute Av2.
Let's check for if v2 is an eigenvector or not.
So we compute this matrix vector product.
That results in this vector right here, and this is interesting.
So I've got minus 15 minus 12 and minus 3 v2 all my entries were originally positive.
If I factor out a negative 3, a negative 3 times 5 is minus 15.
And negative 3 times 4 is a negative 12.
And negative 3 times 1 is a negative 3.
So I've been able to write this as some number times the vector v2.
So this by definition means that v2 is an eigenvector of the matrix A. In fact, we now
know that the eigenvalues associated with that eigenvector is lambda 2 equals negative
3.
So that concludes this short example.
How do you check if a vector is an eigenvector of a matrix?
You simply do the matrix vector multiplication, and see if you can factor that into a fixed
constant scalar times the original vector.
If you can, that factor is an eigenvector.
If you cannot like in this case, then it is not an eigenvector.