If a linear transformation T has matrix A.
If T is a perpendicular projection onto the line y = -5x, then A has:
eigenvector [1, -5] (this is meant to be 2 rows, 1 column) with eigenvalue 1
how would I work out the second eigenvector and eigenvalue?
I thought that because it is a projection, the second eigenvalue would be 0, but does that mean the eigenvector is [0, 0]?
Hint. You're correct about the first eigenvalue and eigenvector, as well as that the second eigenvalue is $0$. Now to find the eigenvector corresponding to the eigenvalue $\lambda=0$, we must find a nonzero vector $v$ such that $Av=\lambda v$, in other words $Av=0$. Thus we just need to find a nonzero vector that is taken to $0$ under the linear transformation. In alternative language, we want to find a vector that spans the nullspace of $A$. Can you continue?