Hermitian Matrix has Real Eigenvalues/Proof 1
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 $\mathbf A^\dagger$ denotes the Hermitian conjugate of $\mathbf A$.
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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:
- $(1): \quad \mathbf v^* \mathbf {A v} = \mathbf v^* \lambda \mathbf v = \lambda \mathbf v^* \mathbf v$
This article, or a section of it, needs explaining. In particular: While $\mathbf A^*$ (since changed to $\mathbf A^\dagger$) was defined as the Hermitian conjugate of $\mathbf A$, that definition applies only to a square matrix. The notation $\mathbf v^*$ is not explained. All subsequent use of $*$ on this page to be reviewed. 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. |
Firstly, note that both $\mathbf v^* \mathbf{A v}$ and $\mathbf v^* \mathbf v$ are $1 \times 1$-matrices.
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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.
$\blacksquare$