# Diagonalization of quadratic form: asymmetric matrix?

I have a quadratic form, say: $A^2+B^2+4AB$.

If I write it in matrix form $V^TQV$ where symmetric $Q=\left[ {\begin{array}{cc} 1 & 2 \\ 2 & 1 \\ \end{array} } \right]$ everything goes fine with diagonalization.

Conversely, if I try to write it as $V^TQ'V$ where non-symmetric $Q'=\left[ {\begin{array}{cc} 1 & 3 \\ 1 & 1 \\ \end{array} } \right]$, I get different eigenvectors that DO NOT diagonalize the form: i.e. when expressing $A$ and $B$ as function of $A'$ and $B'$, the off-diagonal terms do not go away.

What I am asking is the following: since the expression $V^TQ'V$ seems to work perfectly fine in reproducing the form, why I cannot use it for diagonalization?

What is the profound cause of the fact that I need to use a symmetric matrix for diagonalizing a quadratic form?

which are the feature of symmetric matrices that allow diagonalization of quadratic forms?

Symmetric matrices are characterized by the fact that not only are their eigenvalues all real, but they can be written as $S=ODO^T$ where $O$ is orthogonal (with real entries) and $D$ is a diagonal matrix of real eigenvalues. Having such a representation is equivalent to being a symmetric matrix.
• In my example also the eigenevalues of $Q'$ are real, though. There should be more than this, I suspect. – Arnaldo Maccarone Mar 26 '17 at 7:50