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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?


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So, the matrix should be an inverse of itself. – Koba Apr 11 '13 at 16:28
"Distinct" linear factors isn't quite true--take the matrix to be the identity. – user54535 Apr 11 '13 at 21:55
@User24601: Not sure what your mean. The minimal polynomial for the identity is simply $x-1$, and so by itself its a linear factor. Unless you mean to distinguish between "factors" and "factor". – Shahab Apr 12 '13 at 0:45
Ah, my fault. I mistook "characteristic" for "minimal". – user54535 Apr 12 '13 at 4:24
up vote 19 down vote accepted

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 $(A^2)_{ii} \ne 0$.

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This is indeed simpler. I think the summations run over j though. – Shahab Apr 11 '13 at 16:40
This simple argument works in any ordered field regardless of algebraic closure. As a partial converse, note that $\begin{bmatrix}1&i\\i&-1\end{bmatrix}$ is a counterexample in any field where $-1$ has a square root. – Erick Wong Apr 11 '13 at 16:41
This is also equivalent to saying that if the $i$-th column of $A$ is denoted by $C_i$ then $(A^2)_{i,i}=\|C_i\|^2$ and the norm is zero iff the vector is zero so diagonal terms can't be all zero if the columns of $A$ are not all zero ($A\neq0$) – Dolma Apr 11 '13 at 16:41
@Shahab: Fixed, thanks. – Clive Newstead Apr 11 '13 at 16:50

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 $0$.

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This is almost exactly what I thought of...but I don't know why you are writing $Q^T$. By the definition of diagonalizable it should be $Q^{-1}$. Now it's true that a real symmetric matrix is orthogonally diagonalizable, so what you've written is not wrong...but writing it that way seems to add unnecessary complication. – Pete L. Clark Apr 12 '13 at 1:07
@PeteL.Clark yeah I used that a real symmetric matrix is orthogonal diagonalizable I can change it if you want – Dominic Michaelis Apr 12 '13 at 9:51

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 $A=0$.
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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 $A^2=0$.

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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.

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So uniqueness "proves" $D=0$, hence $S=N$, i.e. $S$ nilpotent. We already knew that. – 1015 Apr 11 '13 at 17:26
@julien No by the Dunford decomposition we know that $S$ has the same eigenvalues of $D$. – user63181 Apr 11 '13 at 17:33
So? I see, so you are using Dunford to prove that the spectrum of $S$ is $\{0\}$ without using the minimal polynomial. That's indeed not a simpler argument. And you still have to use that $S$ is diagonalizable to conclude that $S=0$. But hey, it is good make some advertisement for the Dunford decompsition! – 1015 Apr 11 '13 at 17:43

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