Hermitian Matrix has Real Eigenvalues/Proof 2

Theorem

Every Hermitian matrix has eigenvalues which are all real numbers.

Proof

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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

This article, or a section of it, needs explaining. In particular: The notation $\lambda * v$, $\mathbf A * v$ and $\lambda * \innerprod v v$ are not defined. From the context it appears it probably just means scalar product, but in the context, surrounded as it is by redlinks to all sorts of undefined concepts whose meaning is very far from clear, it needs either to be explained, or (preferably) a more conventional notation is used for scalar product. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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.

$\blacksquare$