Theorem 6.10
Let \(V\) be a finite-dimensional inner product space over \(\mathbf{F}\), and let \(\beta\) be a an orthonormal basis for \(V\). If \(T\) is a linear operation on \(V\), then $$ \begin{align*} [T^*]_{\beta} = [T]^*_{\beta} \end{align*} $$


Proof:

Let \(A = [T^*]_{\beta}\). By the Corollary from Theorem 6.5 we know that \(A_{ij} = \langle T^*(v_j), v_i \rangle\). Therefore,

$$ \begin{align*} \langle x, T(y) \rangle &= \overline{\langle T(y), x \rangle} \\ &= \overline{\langle y, T^*(x) \rangle} \\ &= \langle T^*(x), y \rangle \\ &= \overline{A_{ij}} \\ &= (A^*)_{ij}. \ \blacksquare \end{align*} $$


Theorem 6.10 (Corollary)
Let \(A\) be \(n \times n\) matrix. Then \(L_{A^*} = (L_A)^*\)


Proof:

Let \(\beta\) be the standard basis for \(\mathbf{F}^n\). Then, we know that \([L_A]_{\beta} = A\) by Theorem 2.16. We can apply Theorem 6.10 to see that

$$ \begin{align*} [(L_A)^*]_{\beta} &= [L_A]^*_{\beta} \quad \text{(By Theorem 6.10)} \\ &= A^* \\ &= [L_{A^*}]_{\beta} \end{align*} $$

Therefore \(L_{A^*} = (L_A)^*\) as we wanted to show. \(\ \blacksquare\)


References