Nullity and rank bounds for a nilpotent matrix

Question

Let $A=\mathbb R^{11}\to \mathbb R^{11}$ be a linear transformation such that $A^5=0$ and $A^4\neq 0$. Which of the following is true?
a) $\operatorname{null}A\le7$
b) $2\le\operatorname{null}A$
c) $2\le\operatorname{rk}A\le9$

I don't know how i think about $A$ from this information. I think construct such as example where this hold but I can't construct any such of type example which satisfies $A^5=0$ but $A^4\ne0$.

I have an idea now: consider $T:\mathbb R^{11}\to\mathbb R^{11}$ such that $$T(x_1,x_2,...,x_{11})=(x_2,x_3,x_4,x_5,0,0,..,0)$$

$\Rightarrow T^5(x_1,x_2,...,x_{11})=(0,0,...,0)$. So we can say $\operatorname{null}T\le11-\operatorname{rk}T=7$

I'm thinking in using projections and some kind of rotations. This because you need to reduce the dimension of the image in every iteration. — andrés, Feb 09 '18 at 21:41
ok ; it is a good exemple you constructed, but it won't prove it for all $A$ verifying your conditions (although it is not far from it, in fact). — Netchaiev, Feb 09 '18 at 22:12
@andrés.ahumada: it is a bit like that, known as Jordan decomposition. — Netchaiev, Feb 09 '18 at 22:51

Netchaiev · Accepted Answer · 2018-02-10T11:10:20.830

Only the answer a) is true. In fact, more precisely, the following inequalities are true :

$4\leq \text{Rank}(A)\leq 9 $ and $3\leq\text{Nullity}(A)\leq 7$

To prove it you need to know the Jordan decomposition for a nilpotent linear transformation:

A nilpotent linear transformation of degree $u$ (i.e. $A^u=0$ and $A^{u-1}\neq 0$) is similar to a block diagonal matrix :$$J = \begin{bmatrix} J_{p_1} & \; & \; \\ \; & \ddots & \; \\ \; & \; & J_{p_k}\end{bmatrix}$$ where each block $J_{p_i}$ is a square matrix of size $p_i$ and of the form :$$J_{p_i} = \begin{bmatrix} 0 & 1 & \; & \; \\ \; & 0 & \ddots & \; \\ \; & \; & \ddots & 1 \\ \; & \; & \; & 0 \end{bmatrix}.$$ Where for all $i$, $0\leq p_i \leq u$ and at least one $p_i$ is such that $p_i=u$ ; moreover it is easy to see that $Nullity(A)=k$ (the number of blocs).

Here, with $u=5$ and the condition $p_i\leq u$, the minimum number of blocs is $3$ (since you need at least one bloc of five and a bloc of six is not possible to complete the bloc decomposition, so at least two bloc more are necessary to fulfill the 11 dimensions) : the $Nullity(A)\geq 3$ and, using the theorem $Rank(A) +Nullity(A)=11$ you will have $Rank(A)\leq 9$.

Now, you have at least one bloc of size 5, that is a bloc of rank 4 : the rank of the linear transformation is at least higher than 4, so $Rank(A)\geq 4$, which leads, again from $Rank(A) +Nullity(A)=11$, to $Nullity(A)\leq 9$.

Nullity and rank bounds for a nilpotent matrix

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