Hermitian Matrix has Real Eigenvalues

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Theorem

Every Hermitian matrix has eigenvalues which are all real numbers.


Corollary

Every real symmetric matrix has eigenvalues which are all real numbers.


Proof 1

Let $\mathbf A$ be a Hermitian matrix.

Then, by definition:

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

where $\mathbf A^\dagger$ denotes the Hermitian conjugate of $\mathbf A$.



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

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

By definition of eigenvector:

$\mathbf{A v} = \lambda \mathbf v$

Left-multiplying both sides by $\mathbf v^*$, we obtain:

$(1): \quad \mathbf v^* \mathbf {A v} = \mathbf v^* \lambda \mathbf v = \lambda \mathbf v^* \mathbf v$




Firstly, note that both $\mathbf v^* \mathbf{A v}$ and $\mathbf v^* \mathbf v$ are $1 \times 1$-matrices.



Now observe that, using Conjugate Transpose of Matrix Product: General Case:

$\paren {\mathbf v^* \mathbf{A v} }^\dagger = \mathbf v^* \mathbf A^* \paren {\mathbf v^*}^*$

As $\mathbf A$ is Hermitian, and $\paren {\mathbf v^*}^* = \mathbf v$ by Conjugate Transpose is Involution, it follows that:

$\mathbf v^* \mathbf A^\dagger \paren {\mathbf v^*}^* = \mathbf v^* \mathbf{A v}$

That is, $\mathbf v^* \mathbf {A v}$ is also Hermitian.


By Product with Conjugate Transpose Matrix is Hermitian, $\mathbf v^* \mathbf v$ is Hermitian.

So both $\mathbf v^* \mathbf {A v}$ and $\mathbf v^* \mathbf v$ are Hermitian $1 \times 1$ matrices.


Now suppose that we have for some $a,b \in \C$:

$\mathbf v^* \mathbf {A v} = \begin {bmatrix} a \end {bmatrix}$
$\mathbf v^* \mathbf v = \begin {bmatrix} b \end {bmatrix}$

Note that $b \ne 0$ as an eigenvector is by definition non-zero.

By definition of Hermitian matrix:

$\begin {bmatrix} a \end {bmatrix} = \begin {bmatrix} a \end {bmatrix}^*$ and $\begin {bmatrix} b \end {bmatrix} = \begin {bmatrix} b \end {bmatrix}^*$

By definition of Hermitian conjugate:

$\begin {bmatrix} a \end {bmatrix}^* = \begin {bmatrix} \bar a \end {bmatrix}$ and $\begin {bmatrix} b \end {bmatrix}^* = \begin {bmatrix} \bar b \end {bmatrix}$

where $\bar a$ denotes the complex conjugate of $a$.

So by definition of equality of matrices:

$a = \bar a$ and $b = \bar b$

By Complex Number equals Conjugate iff Wholly Real:

$a, b \in \R$, that is, are real.


From equation $(1)$, it follows that:

$\begin {bmatrix} a \end{bmatrix} = \lambda \begin{bmatrix} b \end{bmatrix}$.

Thus:

$a = \lambda b$

Hence because $b \ne 0$:

$\lambda = \dfrac a b$

Hence $\lambda$, being a quotient of real numbers, is real.

$\blacksquare$


Proof 2

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.

$\blacksquare$