Real Symmetric Matrix is Hermitian

Theorem

Every real symmetric matrix is Hermitian.

Proof

Let $\mathbf A$ be a real symmetric matrix.

Then we have:

\(\ds \sqbrk {\mathbf A}^\dagger_{i j}\)

\(=\)

\(\ds \overline {\sqbrk {\mathbf A}_{ji} }\)

Definition of Hermitian Conjugate

\(\ds \)

\(=\)

\(\ds \sqbrk {\mathbf A}_{ji}\)

Complex Number equals Conjugate iff Wholly Real: $\mathbf A$ is Real

\(\ds \)

\(=\)

\(\ds \sqbrk {\mathbf A}_{ij}\)

$\mathbf A$ is Symmetric

So:

$\mathbf A = \mathbf A^\dagger$

Thus, by definition, $\mathbf A$ is Hermitian.

$\blacksquare$

Sources

• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Hermitian conjugate