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# 11.1: Self-adjoint or hermitian operators

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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&-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

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