A symmetric matrix whose square is zero

I was once asked in an oral exam whether there can be a symmetric non zero matrix whose square is zero. After some thought I replied that there couldn't be because the minimal polynomial of such a matrix is guaranteed to be $x^2$ which shows that it isn't diagonalizable. I had to further clarify that a matrix is diagonalizable iff its minimal polynomial is a product of distinct linear factors, and that every symmetric matrix is diagonalizable.

While all this is correct, the examiner mentioned that there is a simpler argument possible but he didn't elaborate on it. I have since been wondering what that simpler argument could be. Can someone give a simpler proof?

Thanks

The $(i,i)^{\text{th}}$ component of the square of an $n \times n$ symmetric matrix $A=(a_{ij})$ is given by $$\sum_{j=1}^n a_{ij}a_{ji} = \sum_{j=1}^n a_{ij}^2$$ If $A \ne 0$ then some $a_{ij} \ne 0$, and then $(A^2)_{ii} \ne 0$.

As Pete said I used that a symmetric square matrix is orthogonal diagonalizable

A symmetric square matrix is diagonalizable hence $$A=Q^{-1} D Q$$ Hence $$A^2 =Q^{-1} D\cdot D Q $$ We multiply with $Q^T$ from right and with $Q$ from left $$Q A^2 Q^{-1} = D^2 $$ As $A^2=0$ we have $$0=D^2$$ So the square of every eigenvalue is $0$ hence all eigenvalues are $0$ hence $A$ must be $0$.

This really depends on the underlying field.

• As Erick Wong has pointed out in a comment to another answer here, there exist complex symmetric matrices whose squares are zero. The example he gave is $\pmatrix{1&i\\ i&-1}$.
• If the underlying field is $GF(2)$, we have $\pmatrix{1&1\\ 1&1}^2=0$.
• If you are talking about real symmetric matrices, then $A^2=0\Rightarrow A=0$. Many answers here have explained why this is the case. Here I will add another one: if $A^2=0$, then $0=x^TA^2x=x^TA^TAx=\|Ax\|^2$ for all vector $x$, i.e. $Ax=0$ for all $x$. Hence $A=0$.

Denote $A^*$ the adjoint, which is simply the transpose in the real case, as I assume it is the case here. So your assumption is $A^*=A$. Then take the trace:

$$0=\mbox{Trace}(A^2)=\mbox{Trace}(A^*A)=\sum |a_{ij}|^2\quad\Rightarrow\quad a_{ij}=0\quad\forall i,j.$$

And this works more generally for $A$ hermitian such that $A^2=0$.

I give an answer but I'm not sure that it'll be a simpler argument:

By the Dunford decomposition we know that $S$ can be written $$S=D+N$$ where $D$ is diagonalizable matrix and $N$ is nilpotent matrix and we have unicity of decomposition but since $S=0+S$ then $D$ must be $0$, hence $S=0$ is the only symmetric matrix that verify the hypothesis.