Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

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 \(\lambda=1,4\).