# Let $A$ be a $5 \times 5$ matrix such that $A^2=0$. Then how to compute the maximum rank for such A?

Attempt : Suppose $A$ has a non-zero eigenvalue $\lambda$. Then corresponding to it's non-zero eigen vector $X$, we have $AX=\lambda X \Rightarrow A^2X=\lambda^2 X\Rightarrow 0=\lambda^2 X$. Which is a contradiction. Hence $\lambda=0$ with algebraic multiplicity $5$.

Looking at all the Jordan normal forms $J$ of $A$, I found one $J$ which gives $J^2=0$ whose $\text {Rank} J=2$. (There is other Jordan normal form with rank$=1$)

Here is that J := $$J = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$

I found that Jordan normal forms with ranks $3,4$ don't satisfy $J^2=0$. So the least upper bound for the rank is $2$. Is my attempt correct? Thanks.

• The minimal polynomial of the matrix divides $x^2$. This gives you an upper bound on the size of the Jordan blocks. – amd May 5 '17 at 7:25
• In your title, use "maximum" instead of "least upper bound": this should be reserved to quantities that vary in $\mathbb{R}$, not in $\mathbb{N}$. – Jean Marie May 5 '17 at 7:30
• @amd True. I know this i.e. Largest size of Jordan block of $\lambda_i =$ degree of factor $(x-\lambda_i)$ in minimal polynomial. So by this way I have two Jordan forms, one with rank $1$, and second with rank $2$. Awesome +1. – Error 404 May 5 '17 at 8:09
• @JeanMarie I see. I am changing it to "maximum". I think for closed and bounded set, maximum=supremum=least upper bound. Here in this case, we have at the max 6 different ranks viz., $\{0,1,2,3,4,5\}$. So we should be able to use both words maximum or supremum. – Error 404 May 5 '17 at 8:13
• Yes of course, but in general, in mathematics, we try to work at "minimum conceptual energy": it's better not to use an involved concept when not necessary. – Jean Marie May 5 '17 at 8:17

Here is a proof not relying on Jordan normal form.

You have exhibited a $5 \times 5$ matrix that has rank $2$.

Let us work by contradiction for the cases $n \geq 3$.

Let us assume that there exists an $A$ with

• (1) rank$(A)\geq 3$ and

• (2) $\ A^2=0.$

By definition of rank function: (1) means that dim$(Im(A)) \geq 3$;

thus, by rank-nullity theorem : dim$(Ker(A)) \leq 5-3=2.$

Let us take any $X \in Im(A)$, it is thus of the form $X=AY$ for a certain $Y$.

Using (2) : $AX=A^2Y=0$ ; thus $X \in Ker(A)$.

As a conclusion: $Im(A) \subset Ker(A)$, which is impossible considering the dimensionalities $\geq 3$ and $\leq 2$ obtained above.

• This is really cool!! Wow. +1. – Error 404 May 5 '17 at 8:23
• It works also nicely without contradiction: $A^2=0$ $\Rightarrow$ $\text{im}\,A\subset\ker A$ $\Rightarrow$ rank$\,A\le\frac{n}{2}$. – A.Γ. May 5 '17 at 13:11
• @ A.Γ. you are right, thanks for the comment! – Jean Marie May 5 '17 at 13:16