Linear Maps and Eigenvalues/Eigenvectors

[joypav]

A problem from matrix analysis. This proof seems to simple... is it correct?

Problem 1:
Suppose $T: V \rightarrow V$ is linear and $\mu \in F$. Show that if there exists a positive integer p such that $ker((T - \mu I)^p)$ $\ne \left\{ 0 \right\}$, then $\mu$ is an eigenvalue of T.

Proof:
$ker((T - \mu I)^p) \ne \left\{ 0 \right\} \implies \exists y \in ker((T - \mu I)^p), y \ne 0$
$\implies (T - \mu I)^p y = 0$

Consider,
$0 \ne y, (T - \mu I)y, (T - \mu I)^2 y , ... , (T - \mu I)^p y = 0$

Then, $\exists j, 0 < j \leq p $, the smallest positive integer so that
$(T - \mu I)^{j-1} y \ne 0$ and $(T - \mu I)^j y = 0$

Let $z=(T - \mu I)^{j-1} y$.
Then $z \ne 0$ and $(T - \mu I) z = (T - \mu I)^j y = 0$
$\implies (T - \mu I) z = 0 $
$\implies \mu$ is an eigenvalue of V.Additionally,
Problem 2:
Suppose $T: V \rightarrow V$ is linear, $\mu \in F$, $x \in V, x \ne 0$, and $(T - \mu I)^p x = 0$ for some $p \in \Bbb{N}$. Show that one of the vectors
$x, (T - \mu I) x, (T - \mu I)^2 x, ... ,(T - \mu I)^{p-1} x$
is an eigenvector of T.

Can this problem be approached similarly?

 

[GJA]

Hi joypav,

Everything for Problem 1 looks good, nicely done. You're correct about Problem 2 as well; it can be done in a similar fashion to Problem 1.