Is the zero matrix the only symmetric, nilpotent matrix with real values?

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?

Solution 1:

Hint: Real symmetric matrices are (orthogonally) diagonalisable. And all eigenvalues of nilpotent matrices are zero.

Solution 2:

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 $A^k=0$.