True or False: If $A^2=0$ for a $10$ by $10$ matrix $A$, then the inequality rank($A$) $\le 5$ must hold.
I am guessing that this is false. Apparently proving or disproving this uses knowledge of basis, change of coordinates, or dimension (the chapter of the problem is about these). I tried proof by contradiction, which didn't work so far.
Hint
Let $Im(A)$ be the image of $A$, so that $rank(A)=\text{dim}(Im(A))$.
Since $A^2 = 0$, $Im(A) \subset Ker(A)$. Now use this and the rank-nullity theorem.
As stated above in a detailed manner $A^2 = 0$ implies $ Im(A) \subset \ker(A) \Rightarrow \dim( Im(A)) \subset \dim(\ker(A)) $.
Now by rank-nullity theorem, $ dim(\ker(A)) + \dim(im(A)) = 10 \Rightarrow 2(dim(im(A)) \le 10 \Rightarrow \dim(im(A)) \le 5 $
For a matrix with $A^2=0$ assuming it be different from null matrix because for the null matrix the case the result is trivial.
Consider $A^2=0$ , $m_A(x)=x^2$
There exists a Jordan block corresponding to the eigenvale $0$ of size $2×2$. And every jordan block corresponding to the eigenvalue $0$ of size $n×n$ has rank $n-1$. So in order to maximize your rank of the matrix you can have jordan blocks of size $2×2$ and not more than this size due to the restriction by the minimal polynomial. So you can have $5$ blocks ,each increasing the rank of the matrix by $1$.so maximum rank can be $5$
In general a matrix $A$ of size $n×n$ such that $A^2=0$ has rank atmost $\lfloor \frac{n}{2}\rfloor$