0
$\begingroup$

A square matrix $A$ is such that $\mathrm{rank}(A^{k})=\mathrm{rank}(A^{k+1})$. Then $\mathrm{rank}(A^{k})=\mathrm{rank}(A^{k+i})$ for all $i$.

I think we prove this by induction on $i$.

Let $i = 2$. We show that rank $(A^{k})$ = rank $(A^{k+2})$.

Now, we know that rank $(A^{k+2})$ = rank $(A^{k+1}A)$ $\le$ min (rank$(A^{k+1})$, rank$(A)$) $\le$ rank $(A^{k+1})$ = rank $(A^{k})$.

However, I cannot proceed further.

$\endgroup$
2
$\begingroup$

Think geometrically. We have $A^{k+1}=A^kA$, so the image of $A^{k+1}$ is contained in the image of $A^k$. If the rank of the two are equal, that means the image is the same. Since $A^{k+1}=AA^k$, this means that $A$ is a bijection on said image.

$\endgroup$
1
$\begingroup$

This can be nicely shown by thinking of $A$ as a linear transformation $f \colon K^n \to K^n$ and looking at the decreasing chain $\operatorname{im} f^p \supseteq \operatorname{im} f^{p+1}$ as follows:

If $A \in \operatorname{Mat}_n(K)$ then let $f \colon K^n \to K^n$ with $f(x) = Ax$ be the associated endomorphism of $K^n$, and let $R_p := \operatorname{im} f^p$ for all $p \geq 0$. Then $$ \operatorname{rank} A^p = \dim \operatorname{im} f^p = \dim R_p \quad \text{for all $p \geq 0$}, $$ as well as $$ f(R_p) = f(\operatorname{im} f^p) = \operatorname{im} f^{p+1} = R_{p+1} \quad \text{for all $p \geq 0$}. $$

Because we have a decreasing chain $$ K^n = R_0 \supseteq R_1 \supseteq R_2 \supseteq R_3 \supseteq \dotsb $$ it follows from $\operatorname{rank} A^{k+1} = \operatorname{rank} A^k$ that $\dim R_{k+1} = \dim R_k$ and thus $R_{k+1} = R_k$.

It then further follows that $$ R_{k+2} = f(R_{k+1}) = f(R_k) = R_{k+1}, $$ and we find inductively that $R_{k+j} = R_{k+j+1}$ for all $j \geq 0$. So we have $$ R_k = R_{k+1} = R_{k+2} = R_{k+3} = \dotsb $$ and thus $R_{k+i} = R_k$ for all $i \geq 0$.

$\endgroup$
-1
$\begingroup$

so you know now that $rank (A^k)=rank (A^{k+1})=rank (A^{k+2})$. you just can take $k'=k+1$ and do the same work for $A^{k'}$. You obtain $$rank (A^k)=rank (A^{k+1})=rank (A^{k'})=rank (A^{k'+1})=rank (A^{k'+2})= rank (A^{(k+1)+2}) = rank (A^{k+3}) $$ and so on

$\endgroup$

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.