prove:
If A is symmetric (with real entries) it is not similar to an antisymmetric matrix
I tried to show this in this way : A is symmetric with real entries so it is a unitary diagonalizable matrix. so $A=PDP^{-1}$ when $D$ is diagonal and $P^{-1}=P^*$ lets assume negatively and get to a contradiction. suppose $B^T=-B$ similar to A . so $A=QBQ^{-1}=A^T=-(Q^{-1})^TBQ^T$
( Q is invertible) I am stuck here , how can I continue in this way ?
A symmetric matrix has only real eigenvalues. An antisymmetric matrix has only purely imaginary eigenvalues. A similarity transformation doesn't change eigenvalues.
So the only way a symmetric matrix can be similar to an antisymmetric matrix is if all eigenvalues are zero. And a symmetric matrix with all eigenvalues equal to zero must be the zero matrix as it is diagonalizable.