# Thread: prove this rank problem

1. ## prove this rank problem

$\alpha : V\rightarrow V$ is an endo. of a finite dimensional vector space V. Show:

$V\geq Im(\alpha)\geq Im(\alpha^{2})\geq ....$

and

$\{0\}\leq Ker(\alpha) \leq Ker(\alpha^{2})\leq ....$

Prove:
$
r(\alpha^{k})-r(\alpha^{k+1})\geq r(\alpha^{k+1})-r(\alpha^{k+2})$

$
r(\alpha^{k})\geq r(\alpha^{k+1})$

2. Originally Posted by silversand

$\alpha : V\rightarrow V$ is an endo. of a finite dimensional vector space V. Show:

$V\geq Im(\alpha)\geq Im(\alpha^{2})\geq ....$
if $v \in \text{Im}(\alpha^{k+1}),$ then $v=\alpha^{k+1}(u)=\alpha^k(\alpha(u)) \in \text{Im}(\alpha^k).$

$\{0\}\leq Ker(\alpha) \leq Ker(\alpha^{2})\leq ....$
if $u \in \ker(\alpha^k),$ then $\alpha^k(u)=0.$ thus: $\alpha^{k+1}(u)=\alpha(\alpha^k(u))=0,$ i.e. $u \in \ker(\alpha^{k+1}).$

$r(\alpha^{k})-r(\alpha^{k+1})\geq r(\alpha^{k+1})-r(\alpha^{k+2})$
for any $i \geq 0$ let $V_i=\text{Im}(\alpha^i).$ so $r(\alpha^i)=\dim V_i.$ from the first part of your problem we know that $V_{i+1}$ is a subspace of $V_i.$ so for any $k \geq 0$ the map $f: \frac{V_k}{V_{k+1}} \longrightarrow \frac{V_{k+1}}{V_{k+2}}$

defined by: $f(u + V_{k+1})=\alpha(u) + V_{k+2}$ is a well-defined onto homomorphism. thus $\frac{V_{k+1}}{V_{k+2}}$ is isomorphic to a quotient of $\frac{V_k}{V_{k+1}}.$ thus: $\dim \left(\frac{V_{k+1}}{V_{k+2}}\right) \leq \dim \left(\frac{V_k}{V_{k+1}}\right).$ hence:

$\dim (V_{k+1}) - \dim(V_{k+2}) = \dim \left(\frac{V_{k+1}}{V_{k+2}}\right) \leq \dim \left(\frac{V_k}{V_{k+1}}\right) = \dim (V_k) - \dim(V_{k+1}). \ \ \Box$

$r(\alpha^{k})\geq r(\alpha^{k+1})$
this is trivial from the first part of your question.