We know that the definition of diagonalization is: for square matrix $A$, to find invertible matrix $P$ and diagonal matrix $\Lambda$ that $$A=P\Lambda P^{-1}$$
If $A$ is symmetric, then above decomposition is always possible: $$A=P\Lambda P^{-1}=P\Lambda P^T$$ where $P$ is also orthogonal. $\Lambda$ is unique up to permutation of its diagonal entries.
My question is: for a symmetric matrix $A$, whether it is possible to find irreversible matrix $Q$ and diagonal matrix $\Sigma$ such that $$A=Q\Sigma Q^T$$ holds? Note here $Q$ is not reversible and $\Sigma$ is different from above $\Lambda$ considering permutation.
Let $A$ be symmetric and not invertible. Let $P$ be orthogonal and $D$ diagonal such that $A=PDP^t$. Then $D$ has at least one zero on its diagonal. Let $J$ be the matrix with zero everywhere $D$ has a zero, and $1$ everywhere $D$ has a nonzero entry. Then $J=J^t$, $JD=DJ=DJD=D$, $A=(PJ)D(JP)^t$, and $PJ$ is not invertible.