Is symmetric matrix over a field F always diagonalizable?

Question

I wonder if the following statement is true or not:

An $n\times n$ matrix $A$ over a field $F$ is diagonalizable if it is symmetric.

If the field $F$ is $\mathbb{R}$ or $\mathbb{C}$, it is true. I wonder if it is true for some other finite fields, such as $\mathbb{Z}_2$.

Your claim about $F=\mathbb C$ is false. It is known that every complex square matrix is similar to some complex symmetric matrix. So, not all complex symmetric matrices are diagonalisable. — user1551, Jan 13 '17 at 19:01

score 10 · Accepted Answer · answered Jan 13 '17 at 17:36

It's not true. For example, the matrix $$ A = \pmatrix{1&1\\1&1} \in \Bbb Z_2^{2 \times 2} $$ fails to be diagonalizable since it is non-zero, but nilpotent. In general: if $x$ is the column-vector of $n$ $1$s, then $xx^T$ fails to be diagonalizable over $\Bbb Z_n$ for the same reason.

Is symmetric matrix over a field F always diagonalizable?

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