Hermitian Matrix has Real Eigenvalues/Proof 2

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Theorem

Every Hermitian matrix has eigenvalues which are all real numbers.


Proof



Let $\mathbf A$ be a Hermitian matrix.

Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates its Hermitian conjugate.


Let $\lambda$ be an eigenvalue of $\mathbf A$.

Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$ of $\mathbf A$.


Denote with $\innerprod \cdot \cdot$ the inner product on $\C$.


\(\ds \lambda * \innerprod v v\) \(=\) \(\ds \innerprod {\lambda * v} v\) Properties of Complex Inner Product
\(\ds \) \(=\) \(\ds \innerprod {\mathbf A * v} v\) Definition of Eigenvector of Linear Operator: $\lambda * v = \mathbf A * v$
\(\ds \) \(=\) \(\ds \innerprod v {\mathbf A^\dagger * v}\) Properties of Adjugate
\(\ds \) \(=\) \(\ds \innerprod v {\mathbf A * v}\) $\mathbf A$ is Hermitian, so $\mathbf A^\dagger = \mathbf A$
\(\ds \) \(=\) \(\ds \innerprod v {\lambda * v}\) Definition of Eigenvector of Linear Operator: $\lambda*v = \mathbf A*v$
\(\ds \) \(=\) \(\ds \overline \lambda * \innerprod v v\) Properties of Complex Inner Product




We have that $v \ne 0$.

Hence by non-negative definiteness of an inner product:

$\innerprod v v \ne 0$

and we can divide both sides by $\innerprod v v$.

Thus:

$\lambda = \overline \lambda$

By Complex Number equals Conjugate iff Wholly Real, $\lambda$ is a real number.

$\lambda$ was arbitrary, so it follows that every eigenvalue is a real number.

Hence the result.

$\blacksquare$