A complex anti/symmetric nilpotent matrix is null?

Question

i know how to prove that

• a real symmetric nilpotent matrix is null
• a real antisymmetric nilpotent matrix is null

But, is a complex symetric nilpotent matrix is null (Q1) ?

I think that a complex antisymetric nilpotent matrix is null. Because if $A$ is complex antisymetric nilpotent matrix, $A$ is diagonalizable, so $A=PDP^{-1}$ with $D$ the diagonal of eigenvalues. But $A$ is nilptotent. For an integer $p$, $A^p=0$ $\Rightarrow D^p=0 \Rightarrow D=0 \Rightarrow A=0$.

Can we prove that a complex antisymetric nilpotent matrix is null without using the argument above (Q2) ?

Answer 1

First, a complex antisymmetric matrix is NOT diagonalizable. Second, a complex antisymmetric nilpotent matrix is not always null.

The following matrix is a counterexample for both: $$ A = \begin{pmatrix} 0 & 1 & 0\\ -1 & 0 & i\\ 0 & -i & 0 \end{pmatrix} $$ $A$ is an antisymmetric matrix, but its characteristic polynomial is $-x^3$ so it is nilpotent and it isn't diagonalizable. $$ B = \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & i\\ 0 & i & 0 \end{pmatrix} $$ Moreover, $B$ is symmetric nilpotent, but not null, with the same reasoning above.

Answer 2

I suppose you mean by symmetric: $A=A^H$, where $A^H$ is taking transpose + complex conjugate. Assume $A= A^H$ nilpotent, $A\ne 0$. Then there is $n\ge 1$ such that $A^{2n}=0$ but $A^n\ne0$. $A^{2n}=0$ implies $$ 0=x^H(A^{2n} x) = (A^nx)^H(A^nx)=\|A^nx\|_2^2 $$ for all $x$, hence $A^n=0$. Contradiction, hence $A=0$. In case $A=-A^H$, you get an additional factor $(-1)^n$ in the above equation, which does not influence the conclusion.

Answer 3

I want to sum up fact on anti/symmetric nilpotent real/complex matrix

1) (A is a real symmetric nilpotent matrix) $\Rightarrow A =0 $

Proof : $A$ is diagonalizable. $(A = PDP^{-1}) \Rightarrow (0=A^p = PD^pD^{-1})$ fon an integer $p$. So $D^p = 0 \Rightarrow D=0 \Rightarrow A=0$.

2) (A is a real antisymmetric nilpotent matrix) $\Rightarrow A =0 $

Proof 1 : $A^2$ is symmetric and nilpotent thus $A^2=0$ (thanks to the previous result). So ${\rm tr}(A^2)=0$ (trace). But $A^2=-(A^T)A$ so $0={\rm tr}(A^2) = -\|A\|^2$ and $A=0$.

(We use the scalar product $(M \mid N) = {\rm tr((M^T)N)}$)

Proof 2 : $A$ is a real antisymmetric matrix so $A$ is diagonalizable in ${\Bbb C}$ (i'm right here ?) We can write the same proof as before for the real symmetric nilpotent matrix.

Proof 3 : let f the endomorphisme of ${\Bbb R}^n$ with matrix $A$. $A$ is nilpotent so $F={\rm Ker}(A)$ is not null. $f(F) \subset F$ and we have the same thing for $F^{\perp}$ (for the usual euclidean scalar product on ${\Bbb R}^n$).

Suppose that $A$ is not null. So $F^{\perp} \neq \{0\}$.

We choose a base according ${\Bbb R}^n = F \oplus F^{\perp}$. In this base, the matrix of $f$ is the block diagonal matrix : $$\begin{pmatrix} 0 & 0 \\ 0 & B \end{pmatrix}$$

$B$ is inversible for ${\rm Ker}(f) \cap F^{\perp}$ is null.

$A$ is nilpotent so $B$ is nilpotent. But $B$ can't be inversible AND nilpotent. So $A=0$.

3) A complex antisymmetric nilpotent is not necessarily null (cf example given by Exodd)

4) A complex symmetric nilpotent is not necessarily null (cf example given by Exodd)

Questions : what are the complex anti / symmetric nilpotent matrix ? Is there any known characterization of these matrix ?