Self-Adjoint Linear Operators

Self-Adjoint Linear Operators

Recall that if $V$ and $W$ are finite-dimensional nonzero inner product space and if $T \in \mathcal L(V, W)$ them the adjoint of $T$ denoted $T^*$ is the linear map $T^* : W \to V$ is defined by considering the linear function $\varphi : V \to \mathbb{F}$ defined by $\varphi (v) = <T(v), w>$ and for a fixed $w \in W$ we define $T^* (w)$ to be the unique vector in $V$ such that $<T(v), w> = <v, T^*(w)>$.

If we now look only at linear operators, say $T \in \mathcal L (V)$ then $T : V \to V$ and $T^* : V \to V$, and in some cases, we will have that $T = T^*$. These type of linear operators are special and are defined below.

Definition: Let $V$ be a finite-dimensional nonzero inner product space. Let $T \in \mathcal L (V)$. Then $T$ is said to be Self-Adjoint if $T = T^*$.

The term "Hermitian" is used interchangeably as opposed to "Self-Adjoint".

We have already seen one type of self-adjoint linear operator, namely the identity operator since $I = I^*$.

For a more complex example, consider the linear operator $T \in \mathcal (\mathbb{F}^2)$ defined by $T(x, y) = (2x + 3y, 3x + 2y)$ and consider the standard basis $\{ (1, 0), (0, 1) \}$ of $\mathbb{F}^2$. Note that $T(1, 0) = (2, 3)$ and $T(0, 1) = (3, 0)$, so we can construct the matrix $\mathcal M (T)$ with respect to the basis $\{ (1, 0), (0, 1) \}$ to be:

(1)
\begin{align} \quad \mathcal M (T, \{ (1, 0), (0, 1) \} = \begin{bmatrix} 2 & 3\\ 3 & 2 \end{bmatrix} \end{align}

As we saw on The Matrix of the Adjoint of a Linear Map, the matrix of $T^*$ with respect to this basis $\{ (1, 0), (0, 1) \}$ can be obtained taking the conjugate transpose of $\mathcal M (T, \{ (1, 0), (0, 1) \}$, however, note that $\mathcal M (T, \{ (1, 0), (0, 1) \} = \mathcal M (T^*, \{ (1, 0), (0, 1) \}$ so $T$ is self-adjoint.

We will now look at some basic properties of self-adjoint matrices.

Proposition 1: Let $V$ be a finite-dimensional nonzero inner product space and let $S, T \in \mathcal L (V)$. If $S$ and $T$ are self-adjoint then $(S + T)$ is also self-adjoint.
  • Proof: Let $S$ and $T$ both be self-adjoint. Then $S = S^*$ and $T = T^*$. Thus we have that:
(2)
\begin{align} \quad S + T = S^* + T^* = (S + T)^* \end{align}
  • Therefore $(S + T) = (S + T)^*$ so $(S + T)$ is self-adjoint.
Proposition 2: Let $V$ be a finite-dimensional nonzero inner product space and let $T \in \mathcal L(V)$ and $a \in \mathbb{R}$. If $T$ is self-adjoint then $aT$ is also self-adjoint.

Note that proposition 2 holds only if $a$ is a real number since in the proof below we require that $a = \bar{a}$ which holds if and only if $a \in \mathbb{R}$.

  • Proof: Let $T$ be self-adjoint and let $a \in \mathbb{R}$. Then $T = T^*$. Thus we have that:
(3)
\begin{align} \quad aT = aT^* = \bar{a}T^* = (aT)^* \end{align}
  • Therefore $(aT) = (aT)^*$ so $aT$ is self-adjoint.

Example 1

Let $T$ be a linear operator on the inner product space $V$. Prove that $(T - T^*)^2$ is self-adjoint.

To prove that $(T - T^*)^2$ is self-adjoint we must show that $(T - T^*)^2 = ((T - T^*)^2)^*$. We have that:

(4)
\begin{align} \quad ((T - T^*)^2)^* = (T - T^*)^*(T - T^*)^* = (T^* - (T^*)^*)(T^* - (T^*)^*) \\ \quad = (T^* - T)(T^* - T) = (-I)(T^* - T)(-I)(T^* - T) = (T - T^*)(T - T^*) = (T - T^*)^2 \end{align}

So indeed $(T - T^*)^2$ is self-adjoint.

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