Consider first the case where \(\Mtrx{A}\) is orthogonally diagonalizable. This means that there is a diagonal matrix \(\Mtrx{D}\) and an orthogonal matrix \(\Mtrx{C}\) such that

\[\Mtrx{A} = \Mtrx{C}\cdot \Mtrx{D}\cdot \MtrxTrnsps{C}\]

But then we find:

\(\MtrxTrnsps{A}\)	\(=\)	\(\MtrxTrnsps{ \left( C\cdot D\cdot \MtrxTrnsps{C} \right) }\)
\(\)	\(=\)	\(\MtrxTrnsps{\left(\MtrxTrnsps{C}\right)}\cdot \MtrxTrnsps{D}\cdot \MtrxTrnsps{C}\)
\(\)	\(= \)	\(\Mtrx{C}\cdot \Mtrx{D}\cdot \MtrxTrnsps{C}\)
\(\)	\(=\)	\(\Mtrx{A}\)

This means that \(\Mtrx{A}\) is symmetric.

What we just showed means that only a symmetric matrix has a chance of being orthogonally diagonalizable. Much harder to prove is that, if \(\Mtrx{A}\) is symmetric, then it is orthogonally diagonalizable. We will not prove this here.