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If a have a matrix $C \in M_n(\mathbb{R})$ with $C^2=O_n$ and $n=2k+1$ that has at least to eigenvalues equal to $0$. Can I say from this that the $rank C\leq \frac{n-1}{2}?$

mathlearning
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    You need only assume $C^2 = O$, then in fact all eigenvalues are $0$, since $C$ is nilpotent. – Minus One-Twelfth Mar 17 '19 at 12:46
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    Since $C^2=0$, then $\operatorname{Im}(C)\subseteq\operatorname{Ker}(C)$. Then $n=\operatorname{dim}\operatorname{Ker}(C)+\operatorname{dim}\operatorname{Im}(C)\geq 2\operatorname{dim}\operatorname{Im}(C)$. Since $n$ is odd and $\operatorname{dim}\operatorname{Im}(C)$ a natural number, then $\frac{n-1}{2}\geq \operatorname{dim}\operatorname{Im}(C)$. – user647486 Mar 17 '19 at 12:56

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$\newcommand{\rank}{\operatorname{rank}}\newcommand{\nullity}{\operatorname{nullity}}\newcommand{\c}{\mathbf{c}}\newcommand{\0}{\mathbf{0}}$Yes. Hints: You want to show that $\rank(C)\le k$. Let the columns of $C$ be $\c_1,\ldots,\c_n$. Then since $C^2 = O$, we have $C\c_j = \0$ for all $j=1,\ldots, n$. Now suppose by way of contradiction that $\rank(C) > k$. Then since (by the rank-nullity theorem) $$\nullity(C) = 2k+1-\rank(C),$$ we have $$\nullity(C) < 2k+1-k = k+1 \Rightarrow \nullity(C) \le k.$$ But the columns of $C$ are all from the null space of $C$ and $\rank(C) > k$. In view of the above, can you explain why this is impossible?

Minus One-Twelfth
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