TITLE: Is the zero matrix the only symmetric, nilpotent matrix with real values?
QUESTION [14 upvotes]: My intuition tells me that the zero matrix is the only matrix that is symmetric and nilpotent with real values, but I'm having trouble proving it (or finding a counterexample.)
I have searched for related problems, but I've found only one where nilpotent was defined as any matrix $A$ where $A^2=0$; using this definition, the problem is pretty easy.  I'm using the more general definition that $A$ is nilpotent if and only if there exists a positive integer $k$ such that $A^k=0$.
Based on my observations while trying to find a counterexample, I've been trying to formulate some argument about the positive semi-definiteness of the entries on the main diagonal, but I'm not getting very far with it.  Is this the right approach?  Is my gut feeling even true?

REPLY [3 votes]: If for $A^2$ the problem is, as you have said, "pretty easy" then for any $A^k$...   
The proof can be made without referring to eigenvalues - it is possible to prove the claim only with the application of basic definitions for symmetric matrix and nilpotent matrix)
Note that for   $A$  which is symmetric matrix   $(A^2)^T= (AA)^T=A^T A^T =(A^T)^2=A^2$ so square of symmetric matrix is also symmetric.  
You have proved, I assume*, that for any non-zero symmetric matrix $A^2 \ne 0 $.     
*(indeed it's easy to prove than  there  are some non-zero values at least on the main diagonal. 
Denote symmetric matrix as matrix consisting of columns
$A=\begin{bmatrix} 
v_{1}  & v_{2}  & \dots & v_{n}  \\  \end{bmatrix}$
Then $A^2=AA=A^TA=\begin{bmatrix} 
v_{1}^T  \\ v_{2}^T  \\ \dots \\ v_{n}^T  \\  \end{bmatrix}\begin{bmatrix} 
v_{1}  & v_{2}  & \dots & v_{n}  \\  \end{bmatrix}= \begin{bmatrix} 
v_{1}^Tv_{1}   & v_{1}^Tv_{2}   & \dots & v_{1}^Tv_{n}    \\ v_{2}^Tv_{1}   & v_{2}^Tv_{2}    & \dots & v_{2}^Tv_{n}     \\\dots  & \dots  &  \dots & \dots  \\ v_{n}^Tv_{1}   & v_{n}^Tv_{2}    & \dots & v_{n}^Tv_{n}   \end{bmatrix} $
but $ v_{i}^Tv_{i} =\Vert v_i \Vert ^2 $ hence on the main diagonal are squares of column vectors lengths. At least one of these vectors $v_i$ is non-zero vector).
If $A^2 \ne 0 $ then the same can be said for $(A^2)^2$, $((A^2)^2)^2$, etc..    
Then always it is possible to find for any $k$ the number $ m > k $  that $A^m \ne 0$ with the application  of iterative squaring the result matrix as above..
But $A^m=A^{m-k}A^k$.
Because  $A^m \ne 0$ so $A^k$ can't be equal $0$, otherwise $A^m$ should have been equal $0$ as well. 
So we have that for any $k<m$ ($m$  can be made arbitrarily big) - it is impossible for non-zero symmetric matrix $A$ to obtain $A^k=0$.