I have a question about the following matrix:
$$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{bmatrix} $$
Find the eigenvalues without calculations and define your answer. Now, I was thinking about this problem. And I thought, yeah ok if you try the vector (1,1,1), you can find 6 as one eigenvalue (and I know you have a double multiplicity 0 too). But than you are doing sort of guessing/calculation work.
I see that the columns are linearly dependant. So I know the dimension of the column space and of the null space.
Thank you in advance.
EDIT: follow up question:
Ok, so you find that the dimension of the null space is 2, so there are 2 eigenvectors when the eigenvalue is 0. Now my question is, can the dimension of the eigenspace be bigger than the amount of eigenvalues? I guess not. I know it can be smaller