Question

matrices: to find an eigenvector, are we allowed to simplify the matrix?

Answer 1

To find an eigenvector you will need to find your eigenvalues first. Simplyfying the matrix function isnt nessasary

Answer 2

You can rearrange it as much as you want (linear operations), it will boil down to the same eigenvector anyway.

Answer 3

then it is (lamda * I - A)v = 0

Answer 4

if it helps, the matrix in question is\[\left[\begin{matrix}4 & -1 & 6 \\ 2 & 1 & 6\\2 & -1 & 8\end{matrix}\right]\]

Answer 5

Its been awhile but im mostly sure that is the correct formula. v is your eiginvector btw. You will need to solve for it

Answer 6

To put is this way. I dont think you can simplfy it. In fact im almost sure you cant but it doesnt matter because you wont have to.

Answer 7

Walleye, I believe lambda * I and A are swapped, but that's it. :)

Answer 8

You could be right. I have a linear algebra book around here. Let me take a look

Answer 9

doesn't matter

Answer 10

i'm just trying to see if i can save some work haha, it's just a lot of FOILing and adding that I don't want to make a mistake in my math

Answer 11

Sometimes you can stare at the matrix a bit and guess eigenvectors and values. For example, notice that the sum of every row is 9. So it would make sense that an eigenvalue would be 9, and its eigenvector would be (1,1,1)

Answer 12

Also, the trace of the matrix (sum of main diagonal elements) helps, because it gives the total of the eigenvalues. In your case, it is 13.
The determinant of the matrix gives the product of the eivenvalues. In the given case, the determinant is 36.
Together with joemath's eigenvalue of 9, you get 9,2,2 as eigenvalues.