Let $N$ be a 9 $\times$ 9 matrix for which $N^3 = 0$. Suppose that $N^2$ has rank 3. Prove that $N$ has rank 6.
I probably need to use the Rank-nullity theorem; dim ker $f +$ rk $f = 9$, and show that dim ker $f = 3$?
I understand that ker $f \subset$ ker $f^2 \subset$ ker $f^3$, so dim ker $f \leq$ dim ker $f^2 = 6$.
How do I proceed?
Best Answer
Let x be the nullity of N. Then, N$^{2}$ has nullity that is between x and 2x. We have that nullity of N$^{2}$ is 6, because its rank is 3.
So, x$\leq$6$\leq$2x.
By the same logic, x$\leq$9$\leq$3x for N$^{3}$. Adding these two inequalities gives,
2x$\leq$15$\leq$5x, Which implies that either 3x = 15 or 5x =15, because 15 is multiple of 3 or 5 and x itself can't be 15.
Now, if 3x = 15 then x=5, which can't be true because N has rank 6 and it would be exceed the dimension of vector space.
So, x=3. Then the rank of N is 6.