When $e$ is an eigenvector to $G$ prove that $e$ is an eigenvector of $G+k I$ and $G^2$

Question

I've just started learning matrices and I've been shown how to perform row operations, how to find an inverse matrix, how to find eigenvectors from a given matrix, reduced-echelon form, basis, dim(M), rank(M), etc. However, I've just come across this question:

The vector e is an eigenvector of the square matrix G. Show that:
$i.$ e is an eigenvector of G + kI, where k is a scalar and I is an identity matrix.
$ii.$ e is an eigenvector of G$^2$

I know that applying the general rule $Ax = \lambda x \implies (A-\lambda I)x = 0$ will somehow help me here. But without being given any matrix to work with I'm not sure what sort of result I'm meant to achieve, or how I'm supposed to "prove" the two statements.

How should this be done?

Dominic Michaelis · Accepted Answer · 2013-03-14T08:19:55.257

2

Just use the definition of Eigenvectors, let's say $\lambda$ is the eigenvalue, so that $Ge=\lambda e$ $$(G+k I) \cdot e= G e + k I e = \lambda e + k e = (\lambda +k ) e$$ for the second $$G^2 e = G G e = G (G e) = G \lambda e= \lambda G e = \lambda^2 e$$

edited Mar 14 '13 at 08:19

answered Mar 14 '13 at 07:27

Dominic Michaelis

19,935

Is that first line the definition of eigenvectors? Please explain. It's slightly different from the equation I know of, so I'm not sure where it comes from. – Ozzy Mar 14 '13 at 07:35
A vector $x\neq 0$ is called an eigenvector to the eigenvalue $\lambda$ if $A x = \lambda x$ – Dominic Michaelis Mar 14 '13 at 07:43
Sorry, I don't get why $(G+kI)\cdot e = \lambda e + ke$. All I have so far is that $Ax = \lambda Ix \implies A=\lambda I$; $x$ is an eigenvector. Could you shed some more light on that please? – Ozzy Mar 14 '13 at 08:12
you can't devide by $x$ they are vectors, you have the equation $(A-\lambda I)x= Ax-\lambda x =0\implies Ax=\lambda x$ and i just multiply it out – Dominic Michaelis Mar 14 '13 at 08:17
Right, but I still don't understand the connection between $Gx = kx$ and $(G+kI)\cdot x = \lambda x + kx$ :S – Ozzy Mar 14 '13 at 08:21
Take $e=x$ ?! multiply it out, use that the identity is the identity sry i really don't know where the proble mis, maybe its better to come in the chat – Dominic Michaelis Mar 14 '13 at 08:25
let us continue this discussion in chat – Ozzy Mar 14 '13 at 08:26

When $e$ is an eigenvector to $G$ prove that $e$ is an eigenvector of $G+k I$ and $G^2$

1 Answers1