- #1

joypav

- 151

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**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?