Let $M$ be $10\times10$ matrix such that $M^{4}=0$ and $M^{i} \neq 0$ for $i=1,2,3$. Then what are the minimum and maximum ranks of the matrix. Where minimum rank and maximum rank is r and R respectively ?
a) R= 7, r=3
b)R= 6, r=4
c)R= 6, r=3
d)R= 7, r=4
think since there $M^{i}$ is not equal to $0$ for three values of $i$ so the minimum rank is $3$ But how to find the maximum rank? I have not studied it's not part of curriculum that I have to solve without Jordan form.
Your answer for the minimal rank is correct.
I find it easiest to think of this question in terms of the Jordan form of $M$. Because Because $M$ has minimal polynomial $p(x) = x^4$, we know that $M$ has $0$ as its only eigenvalue, and that the largest Jordan block has size $4$. On the other hand, $\dim \ker(A) = 10 - \operatorname{rank}(A)$ is the total number of Jordan blocks associated with zero.
Thus, our question amounts to finding the Jordan form matrix (with only zero-blocks) that contains the fewest blocks, each block of size at most $4$. Since $M$ must contain at least $10/4 = 2.5$ such blocks, we conclude that the kernel has dimension at least $3$, which is to say that the rank of $M$ is at most $7$.