Properties of Adjoints of Linear Maps

Recall from The Adjoint of a Linear Map page that if $V$ and $W$ are finite-dimensional nonzero inner product spaces and that $T \in \mathcal L (V, W)$ then the adjoint of $T$ is the linear map $T^* \in \mathcal L (W, V)$ 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 $T^*$ to be the unique vector such that $< T(v), w> = <v, T^*(w)>$.

We will now look at some properties of adjoints in the following propositions.

Proposition 1: Let $V$ and $W$ be finite-dimensional nonzero inner product spaces. Let $S, T \in \mathcal L(V, W)$. Then $(S + T)^* = S^* + T^*$.

• Proof: We note that $<(S + T)(v), w> = <v, (S + T)^*(w)>$ by the definition of an adjoint. Furthermore, we have that:

(1)

\begin{align} \quad <(S + T)(v), w> = <S(v) + T(v), w> = <S(v), w> + <T(v), w>\\ = <v, S^*(w)> + <v, T^*(w)> = <v, S^*(w) + T^*(w)> = <v, (S^* + T^*)(w)> \end{align}

• Thus we have that $(S + T)^* = S^* + T^*$. $\blacksquare$

Proposition 2: Let $V$ and $W$ be finite-dimensional nonzero inner product spaces. Let $T \in \mathcal L(V, W)$ and let $a \in \mathbb{F}$. Then $(aT)^* = \overline{a}T^*$.

• Proof: We note that $<(aT)(v), w> = <v, (aT)^*(w)>$. Furthermore, we have that:

(2)

\begin{align} \quad <(aT)(v), w> = <aT(v), w> = a<T(v), w> = a<v, T^*(w)> = <v, \overline{a}T^*(w)> \end{align}

• Thus we have that $(aT)^*(w) = \overline{a}T^*(w)$. $\blacksquare$

Proposition 3: Let $V$ and $W$ be finite-dimensional nonzero inner product spaces. Let $T \in \mathcal L(V, W)$. Then $(T^*)^* = T$.

• Proof: We note that $<(T^*)^*(v), w> = <v, (T^*)^*(w)>$. Furthermore, we have that:

(3)

\begin{align} \quad <T^*(w), v> = \overline{<v, T^*(w)>} = \overline{<T(v), w>} = <w, T(v)> \end{align}

• Thus we have that $(T^*)^* = T$. $\blacksquare$

Proposition 4: If $I$ is the identity operator on a finite-dimensional nonzero inner product space $V$ then $I^* = I$.

• Proof: We note that $<I(v), w> = <v, I(w)> = <v, w>$. So $I^*(w) = w$ and so $I^* = I$. $\blacksquare$

Proposition 5: Let $U$, $V$ and $W$ be finite-dimensional nonzero inner product spaces. Let $S \in \mathcal L(W, U)$ and let $T \in \mathcal (V, W)$. Then $(ST)^* = T^*S^*$.

• Proof: We note that $<(ST)(v), w> = <v, (ST)^*(w)>$. Furthermore we have that:

(4)

\begin{align} \quad <(ST)(v), w> = <S(T(v)), w> = <T(v), S^*(w)> = <v, T^*(S^*(w))> = <v, (T^*S^*)(w)> \end{align}

• Thus we have that $(ST)^* = T^*S^*$. $\blacksquare$

Example 1

Let $T$ be a linear operator on the inner product space $V$. Prove that $T = T^2$ if and only if $T^* = (T^*)^2$.

$\Rightarrow$ Suppose that $T = T^2$. Then if we take the adjoint of both sides of this equation then we get that:

(5)

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

Here we applied Proposition 5 to the second equality. Therefore $T^* = (T^*)^2$.

$\Leftarrow$ Suppose that $T^* = (T^*)^2$. Then if we take the adjoint of both sides of this equation then we get that:

(6)

\begin{align} \quad (T^*)^* = ((T^*)^2)^* \\ \quad T = (T^* T^*)^* \\ \quad T = ((T^*)^* (T^*)^*) \\ \quad T = (TT) \\ \quad T = T^2 \end{align}

Here we applied Proposition 3 and Proposition 5. Therefore $T = T^2$.