11.1: Self-adjoint or hermitian operators

Let $V$ be a finite-dimensional inner product space over $\mathbb{C}$ with inner product $\inner{\cdot}{\cdot}$. A linear operator $T\in\mathcal{L}(V)$ is uniquely determined by the values of

$$\inner{Tv}{w}, \quad \text{for all \(v,w\in V\).}$$

This means, in particular, that if $T,S\in\mathcal{L}(V)$ and

$$\begin{equation*} \inner{Tv}{w} = \inner{Sv}{w} \quad \text{for all \(v,w \in V\),} \end{equation*}$$

then $T=S$. To see this, take $w$ to be the elements of an orthonormal basis of $V$.

Definition 11.1.1. Given $T\in\mathcal{L}(V)$, the adjoint (a.k.a. hermitian conjugate) of $T$ is defined to be the operator $T^*\in\mathcal{L}(V)$ for which

$$\inner{Tv}{w} = \inner{v}{T^*w}, \quad \text{for all \(v,w\in V\)}$$

Moreover, we call $T$ self-adjoint (a.k.a.hermitian}) if $T=T^*$.

The uniqueness of $T^*$ is clear by the previous observation.

Example 11.1.2. Let $V=\mathbb{C}^3$, and let $T \in \cal{L}(\mathbb{C}^3)$ be defined by $T(z_1,z_2,z_3)=(2z_2+iz_3,iz_1,z_2)$. Then
$$\begin{equation*} \begin{split} \inner{(y_1,y_2,y_3)}{T^*(z_1,z_2,z_3)} &= \inner{T(y_1,y_2,y_3)}{(z_1,z_2,z_3)}\\ &= \inner{(2y_2+iy_3,iy_1,y_2)}{(z_1,z_2,z_3)}\\ &= 2y_2\overline{z_1} + iy_3 \overline{z_1} +iy_1\overline{z_2} + y_2 \overline{z_3}\\ &= \inner{(y_1,y_2,y_3)}{(-iz_2,2z_1+z_3,-iz_1)} \end{split} \end{equation*}$$

so that $T^*(z_1,z_2,z_3)=(-iz_2,2z_1+z_3,-iz_1)$. Writing the matrix for $T$ in terms of the canonical basis, we see that
$$\begin{equation*} M(T) = \begin{bmatrix} 0&2&i\\ i&0&0\\ 0&1&0 \end{bmatrix} \quad \text{and} \quad M(T^*) = \begin{bmatrix} 0&-i&0 \\ 2&0&1 \\ -i&0& 0\end{bmatrix}. \end{equation*}$$

Note that $M(T^*)$ can be obtained from $M(T)$ by taking the complex conjugate of each element and then transposing. This operation is called the conjugate transpose of $M(T)$, and we denote it by $(M(T))^{*}$.

We collect several elementary properties of the adjoint operation into the following proposition. You should provide a proof of these results for your own practice.

Proposition 11.1.3. Let $S,T\in \mathcal{L}(V)$ and $a\in \mathbb{F}$.

1. $(S+T)^* = S^*+T^*$.
2. $(aT)^* = \overline{a} T^*$.
3. $(T^*)^* = T$.
4. $I^* = I$.
5. $(ST)^* = T^* S^*$.
6. $M(T^*) = M(T)^*$.

When $n=1$, note that the conjugate transpose of a $1\times 1$ matrix $A$ is just the complex conjugate of its single entry. Hence, requiring $A$ to be self-adjoint ($A=A^*$) amounts to saying that this sole entry is real. Because of the transpose, though, reality is not the same as self-adjointness when $n > 1$, but the analogy does nonetheless carry over to the eigenvalues of self-adjoint operators.

Proposition 11.1.4. Every eigenvalue of a self-adjoint operator is real.

Proof. Suppose $\lambda\in\mathbb{C}$ is an eigenvalue of $T$ and that $0\neq v\in V$ is a corresponding eigenvector such that $Tv=\lambda v$. Then

$$\begin{equation*} \begin{split} \lambda \norm{v}^2 &= \inner{\lambda v}{v} = \inner{Tv}{v} = \inner{v}{T^*v}\\ &= \inner{v}{Tv} = \inner{v}{\lambda v} = \overline{\lambda} \inner{v}{v} =\overline{\lambda} \norm{v}^2. \end{split} \end{equation*}$$

This implies that $\lambda=\overline{\lambda}$.

Example 11.1.5. The operator $T\in \mathcal{L}(V)$ defined by $T(v) = \begin{bmatrix} 2 & 1+i\\ 1-i & 3 \end{bmatrix} v$ is self-adjoint, and it can be checked (e.g., using the characteristic polynomial) that the eigenvalues of $T$ are $\lambda=1,4$.

Contributors

• Isaiah Lankham, Mathematics Department at UC Davis
• Bruno Nachtergaele, Mathematics Department at UC Davis
• Anne Schilling, Mathematics Department at UC Davis

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