Real Symmetric Matrix is Hermitian
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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