Theorem 6.9
Let \(V\) be a finite-dimensional inner product space over \(\mathbf{F}\), and let \(T\) be a linear operator on \(V\). Then there exists a unique function \(T^*:V \rightarrow V\) such that \( \langle T(x), y \rangle = \langle x, T^*(y) \rangle\) for all \(x,y \in V\). Furthermore \(T^*\) is linear.


Proof:

Let \(y \in V\). Define \(g(x): V \rightarrow \mathbf{F}\) by \(g(x) = \langle T(x), y \rangle\) for all \(x \in V\). \(g\) is linear. To see this, consider \(x_1, x_2 \in V\), then

$$ \begin{align*} g(cx_1 + x_2) &= \langle T(cx_1 + x_2), y \rangle \\ &= \langle T(cx_1) + T(x_2), y \rangle \\ &= \langle cT(x_1), y \rangle + \langle T(x_2), y \rangle \\ &= cg(x_1) + g(x_2) \end{align*} $$

So \(g\) is linear. By Theorem 6.8, there exists a unique vector \(y' \in V\) such that \(g(x) = \langle x, y' \rangle\). That is \(\langle T(x), y \rangle = \langle x, y' \rangle\) for all \(x \in V\). Now, define \(T^*: V \rightarrow\) by \(T^*(y) = y'\) so now we have \(\langle T(x), y \rangle = \langle x, T^*(y) \rangle\). \(T^*\) is also linear, to see this, let \(y_1, y_2 \in V\) and \(c \in \mathbf{F}\). Then for any \(x \in V\),

$$ \begin{align*} \langle x, T^*(cy_1 + y_2) \rangle &= \langle T(x), cy_1 + y_2 \rangle \\ &= \langle T(x), cy_1 \rangle + \langle T(x), y_2 \rangle\\ &= \bar{c}\langle T(x), y_1 \rangle + \langle T(x), y_2 \rangle\\ &= \bar{c}\langle x, T^*(y_1) \rangle + \langle x, T^*(y_2) \rangle \\ &= \langle x, cT^*(y_1) + T^*(y_2) \rangle \end{align*} $$

Since \(x\) is arbitrary, then \(T^*(cy_1 + y_2) = cT^*(y_1) + T^*(y_2)\) by Theorem 6.1(e). Finally we need to show that \(T^*\) is unique. Suppose it wasn’t. Then, let \(U: V \rightarrow V\) be linear such that it satisfies \(\langle T(x), y \rangle = \langle x, U(y) \rangle\) for all \(x, y \in V\). But this means that \(\langle x, U(y) \rangle = \langle x, T^*(y) \rangle\) for all \(x, y \in V\). Therefore, \(T^* = U\). \(\ \blacksquare\)

Note here that

$$ \begin{align*} \langle x, T(y) \rangle &= \overline{\langle T(y), x \rangle} \\ &= \overline{\langle y, T^*(x) \rangle} \\ &= \langle T^*(x), y \rangle \end{align*} $$

So we can shift back and forth between the two.


References