Let $A$ be a $n \times n$ matrix. Show that if $A^k=0$ then $\operatorname{rank} A \leq n(k-1)/k$

Question

I proved it for the case $A^2$ using rank-nullity. I suspect this might be proved using induction.

Are you familiar with Jordan Normal/Canonical Forms? – user757704 Mar 09 '20 at 20:34 — user757704, Mar 09 '20 at 20:34
I think there is a considerably more elementary solution – Mohammed M. Zerrak Mar 09 '20 at 20:37 — Mohammed M. Zerrak, Mar 09 '20 at 20:37

score 5 · Accepted Answer · answered Mar 09 '20 at 20:38

You may prove that, for two $n\times n$ matrices $P, Q$ you have $\text{Nullity}(PQ)\le\text{Nullity}(P)+\text{Nullity}(Q)$. (See e.g. https://math.stackexchange.com/a/2878262/700480 - it is expressed for linear operators in terms of ranks, but easily converted to nullities.)

This makes it easy to prove that $\text{Nullity}(A^k)\le k\cdot\text{Nullity}(A)$ (e.g. induction on $k$).

Thus, if $A^k=0$, you have $n=\text{Nullity}(A^k)\le k\cdot\text{Nullity}(A)=k(n-\text{Rank}(A))$. Now solve this for $\text{Rank}(A)$ to get the required inequality.

Let $A$ be a $n \times n$ matrix. Show that if $A^k=0$ then $\operatorname{rank} A \leq n(k-1)/k$

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