Linear Algebra - help wanted with finding geometric multiplicity and eigenspace of...

Your third line does not follow from your first. Could you clarify that the original matrix is correctly written?

Sorry, typo, I've corrected it now

Geometric multiplicity of an eigenvalue is by definition never 0. Indeed, if the multiplicity were 0, it wouldn't be an eigenvalue, because that would mean the eigenspace had dimension 0, so contained only the point 0: that is, there would be no eigenvector for that eigenvalue.

Can you give me without any thought an eigenvector for that matrix, and hence show that the geometric multiplicity is >= 1? (Hint if necessary: it's already a lower-diagonal matrix.) ETA: by "lower-diagonal" I mean "lower-triangular".

The task is then either to find another eigenvector which is not a multiple of that one, or to show that no such eigenvector exists. There is a system of simultaneous equations to solve to find the eigenvectors. Can you tell me what that system is?

Thank you for your reply

Ok, I now understand that all eigenvalues must have at least one eigenvector, but i still can't see what the eigenvector would be for the eigenvalue of 2.

$\begin{pmatrix} 2 & 0 \\5 & 2 \end{pmatrix}$ $\begin{pmatrix} V_1 \\ V_2 \end{pmatrix} = 2\begin{pmatrix} V_1 \\ V_2 \end{pmatrix}[br]$

So $2V_1 = 2V_1$ which is not very helpful

$5V_1 + 2V_2 = 2V_2$
$5V_1 = 0$
Therefore $V_1 = 0$

So how do I get anything but 0?

You have determined that for any eigenvector of eigenvalue 2 (and hence for any eigenvector of any eigenvalue, since the only eigenvalue is 2), the first entry of that eigenvector must be 0. That is, all eigenvectors must lie on the span of the vector {0, 1}. Does that help?