Scientific claim

Self-Adjoint Operators Have Real Eigenvalues

Claim

Let $V$ be a complex inner-product space, and let $A$ be a linear operator on $V$ .

The operator $A$ is self-adjoint when it equals its adjoint:

A=A^\dagger.

Here, $A^\dagger$ is the unique operator satisfying

\langle u,Av\rangle = \langle A^\dagger u,v\rangle

for every $u,v\in V$ , where $\langle\cdot,\cdot\rangle$ denotes the inner product on $V$ .

Every eigenvalue $\lambda$ of a self-adjoint operator $A$ is real:

\lambda\in\mathbb R.

Here, $\lambda$ is an eigenvalue of $A$ when there exists a nonzero vector $v\in V$ satisfying

Av=\lambda v.

Theoretical: SupportedExperimental: Not applicable

Assumptions

No assumptions recorded.

Proof

A complex square matrix $A$ is self-adjoint when it equals its conjugate transpose:

A=A^\dagger.

The conjugate transpose $A^\dagger$ is obtained by transposing $A$ and taking the complex conjugate of every entry:

A^\dagger=\overline{A}^{\,T}.

Equivalently, the matrix entries satisfy

A_{ij}=A_{ji}^*

for every pair of indices $i$ and $j`.

Every eigenvalue of a self-adjoint matrix is real.

Examples

Consider the matrix
$A= \begin{pmatrix} 2 & 0\\ 0 & 5 \end{pmatrix}.$
Because every entry is real and $A$ equals its transpose,
$A^\dagger=A.$
Therefore, $A$ is self-adjoint.
Its eigenvalues are
$\lambda_1=2, \qquad \lambda_2=5.$
Both eigenvalues are real.
Consider the matrix
$A= \begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}.$
The matrix is real and symmetric, so its conjugate transpose equals itself:
$A^\dagger=A.$
Therefore, $A$ is self-adjoint.
Its eigenvalues are
$\lambda_1=1, \qquad \lambda_2=3.$
Both eigenvalues are real.
Consider the complex matrix
$A= \begin{pmatrix} 2 & i\\ -i & 3 \end{pmatrix},$
where $i$ is the imaginary unit satisfying
$i^2=-1.$
Taking the conjugate transpose gives
$A^\dagger= \begin{pmatrix} 2 & i\\ -i & 3 \end{pmatrix}.$
Therefore,
$A^\dagger=A,$
so $A$ is self-adjoint.
Its eigenvalues are
$\lambda_1=\frac{5-\sqrt{5}}{2}, \qquad \lambda_2=\frac{5+\sqrt{5}}{2}.$
Both eigenvalues are real.
Let $V=\mathbb C$ , where $\mathbb C$ denotes the set of complex numbers.
Define the operator $A$ by
$A(z)=4z$
for every $z\in\mathbb C$ .
Because multiplication by the real number $4$ equals its own adjoint,
$A^\dagger=A.$
Therefore, $A$ is self-adjoint.
Every nonzero $z\in\mathbb C$ satisfies
$A(z)=4z.$
The operator therefore has the real eigenvalue
$\lambda=4.$

Evidence

Determining self-adjointness from the matrixAdam Glesser · 2023-11-17 · supporting

Related Concepts

No dependency connections recorded.