Let $A$ be a symmetric matrix of order $n$ and $A^2=0$ . Is it necessarily true that $A=0$

Question

Let $A$ be a symmetric matrix of order $n$ and $A^2=0$ . Is it necessarily true that $A=0$ .

My approach :

I tried to experiment with some $2\times 2$ matrices but never gotten any far .

Now, Wikipedia says that there exists a diagonal matrix $D$ and orthogonal matrix $Q$ such that $D=Q^t A Q$ . So $D^2=Q^t A^2 Q=0$ . As $D$ is diagonal matrix with real entry we get $D=0$ .

So $A=QDQ^t =0 $ I think my proof is correct . I just want know if there is any way to prove this without citing any big theorems or in more elementary way . The problem is quoted from a part of the web-text that only used elementary definitions like what a symmetric matrix is . So i'm curious if there is an elementary solution to the problem .

Answer 1

An easy and natural way to do this is using inner products. Note that if $A^2v=0$, then $$0=\langle v,A^2v\rangle=\langle A^Tv, Av\rangle=\langle Av,Av\rangle=\|Av\|^2$$ which implies $Av=0$. In particular, if $A^2=0$, we get $Av=0$ for all $v$ and so $A=0$.

If you don't like the language of inner products and want something entirely in the elementary language of matrices, use the definition $\langle v,w\rangle=v^Tw$. So the equation above becomes $$0=v^TA^2v=v^TA^TAv=(Av)^TAv$$ and now you observe that for any vector $w$, $w^Tw$ is just the sum of the squares of the entries of $w$ and so can only be $0$ if $w=0$.

Answer 2

If $A \neq 0$, then there are some $1 \leq i,j \leq n$ ensuring that if $a_{ij}$ is the $(i,j)$ element of $A$ then $a_{ij} \neq 0$. The matrix $A$ is by assumption symmetric; so $a_{ij} = a_{ji}$. If $b_{ij}$ is the $(i,j)$ element of $A^{2}$, then $b_{ij} \geq a_{ij}^{2} > 0$.