0
$\begingroup$

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?

  • 2
    Exercise $4.3$ [here](http://pub.math.leidenuniv.nl/~luijkrmvan/linalg2/2016/exerc4.pdf), right? – 2017-01-10
  • 0
    $N^3={N^2}{N}=0$ – 2017-01-10
  • 0
    Yes @DietrichBurde – 2017-01-10

1 Answers 1

2

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.