If $A$ is a $n\times n$ matrix with entries from $\mathbb R$ or $\mathbb C$ (complex) and ${A^{T}}A=0$ prove $\operatorname{Rank}(A)\le \frac{n}{2}$

Question

If $A$ is a $n\times n$ matrix with entries from $\mathbb R$ or $\mathbb C$ (complex) and ${A^{T}}A=0$ prove that $$\operatorname{Rank}(A)\le \frac{n}{2}.$$

i know how to prove it when the entries are Real numbers but i have trouble when the entries are complex numbers.

score 1 · Answer 1 · edited Feb 08 '21 at 22:31

We can use Sylvester rank inequality $$\operatorname{rank}(A) + \operatorname{rank}(B) \le \operatorname{rank}(AB) + n$$ which is valid for $n\times n$ matrices $A,B$ over any field to obtain $$2\operatorname{rank}(A)=\operatorname{rank}(A^T) + \operatorname{rank}(A)\le\operatorname{rank}(A^TA) + n = n.$$

In case of real numbers from $A^TA = 0$ we actually get $A=0$ since $\operatorname{Tr}(A^TA)$ is the sum of squares of all elements of $A$.

If $A$ is a $n\times n$ matrix with entries from $\mathbb R$ or $\mathbb C$ (complex) and ${A^{T}}A=0$ prove $\operatorname{Rank}(A)\le \frac{n}{2}$

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