The answer choices are:
- $\operatorname{rank}(A) = 0$
- $\operatorname{rank}(A) \leq \frac{n}{2}$ (and possibly $>0$)
- $\operatorname{rank}(A) \leq n-1$ (and possibly $> \frac{n}{2}$)
I know that 1. is false, but I don't know about 2 and 3.
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Arthur
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user469779
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1the second one must be true. – Yanko Aug 05 '17 at 18:08
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1Hint: there are two inequalities that relate the rank of two matrices $B$ and $C$, and the rank of $BC$. – Arthur Aug 05 '17 at 18:10
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@yanko even with the "(and possibly)"? – Arthur Aug 05 '17 at 18:11
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@ Arthur, consider the matrix with (1,0) and (0,0) this is of rank $1$ and it's squared is zero. – Yanko Aug 05 '17 at 18:14
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@yanko I missed $\geq$, thought it was $>$. In fact, checking it, it was $>$ before amWhy changed it. – Arthur Aug 05 '17 at 18:21
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a counter example for $3$ is the 4*4 identity matrix with one of the 1's replaced by zero.
lets prove that $2$ must be true, you know that $$n=dim (Im A) +dim (ker A)$$ let $v\in ImA$ then by assumpation, $v\in Ker(A)$ it follows that $Im A \subseteq Ker (A)$ and therefore $Rank A =dim Im(A) \leq \frac{n}{2}$
Yanko
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A proof for the second statement. We have $A^2 = A[a_1 \ldots a_n] = O$. Take any $x \in \text{coll}(A)$. Then $x = c_1a_1 + \ldots +c_na_n$. But then $$Ax = A(c_1a_1 + \ldots c_na_n) = c_1Aa_1 + \ldots c_nAa_n = 0,$$ hence $x \in \text{null}(A)$ and $\text{nullity}(A) \geqslant \text{rank}(A)$. But then, $$\text{rank}(A) + \text{rank}(A) \leqslant \text{rank}(A) + \text{nullity}(A) = n \Rightarrow rank (A) \leqslant n/2 $$
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1yes this is exactly the same proof as mine with Ker replaced by nullity... – Yanko Aug 05 '17 at 18:18