0
$\begingroup$

I'm studying about linear algebra and came across with the following:

Let $A\in \mathcal{M}_{n\times n}(K)$ for some field $K$. If $\text{dim}\,K^n = n$ is finite then $\text{Im}(A^i)=\text{Im}(A^{i+1})$ for some $i\leq n.$

Why is this true? I get it that $\text{Im}(A^{i+1})\subseteq \text{Im}(A^{i})$, but why does the statement hold for some $i\leq n$? Where does the $i\leq n$ come from?

Please let me know if my question is unclear. Thank you for your help!

$\endgroup$
1
$\begingroup$

Hint: For each $i$, $\dim(\operatorname{Im}(A^i))$ is an integer between $0$ and $n$, and $\dim(\operatorname{Im}(A^{i+1})) \leq \dim(\operatorname{Im}(A^i))$.

$\endgroup$
  • $\begingroup$ Thank you very much! =) Got it! $\endgroup$ – jjepsuomi Feb 6 '16 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.