number of eigenvalues = dimension of eigenspace

Question

Is this true in general? What about: number of negative eigenvalues = dimension of span(eigenectors for the negative eigenvalues)?

Or even more generally: number of eigenvalues greater than 4.3 = dimension of span(eigenvectors for the eigenvalues greater than 4.3)?

Is (number of negative eigenvalues) here distinct eigenvalues or not?

Any help would be much appreciated, as I cna't seem to find a proof anywhere (not even sure if it's true)

Thanks!

Answer 1

Not true. For the matrix \begin{bmatrix} 2 &1\\ 0 &2\\ \end{bmatrix} 2 is an eigenvalue twice, but the dimension of the eigenspace is 1.

Roughly speaking, the phenomenon shown by this example is the worst that can happen. Without changing anything about the eigenstructure, you can put any matrix in Jordan normal form by basis-changes. JNF is basically diagonal (so the eigenspaces are exactly the coordinate axes/planes/whatever), except it can contain blocks like this one, with a repeated evalue on the diagonal, and 1s above the diagonal.