What is Adjoint; symmetric endomorphisms? — Mathematics

Definition 12.11 University Mathematics — Year 2 · Chapter 12 — Quadratic Forms

For $u \in \mathcal{L}(E)$ , the adjoint $u^*$ is the unique endomorphism with

\langle u(x), y\rangle = \langle x, u^*(y)\rangle \qquad (x, y \in E);

in an orthonormal basis, $\operatorname{Mat}(u^*) = \operatorname{Mat}(u)^{\mathsf T}$ . $u$ is symmetric (self-adjoint) when $u^* = u$ — equivalently its matrix in an orthonormal basis is symmetric.

Examples

Example 12.12 (The adjoint depends on the inner product)

On $\R^2$ take the weighted inner product $\langle x, y\rangle_D = x_1y_1 + 2x_2y_2$ (matrix $D = \operatorname{diag}(1,2)$ ) and $u$ of matrix $A = \begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$ in the canonical basis. From $\langle u(x), y\rangle_D = (Ax)^{\mathsf T}Dy = x^{\mathsf T}(A^{\mathsf T}D)y$ and $\langle x, u^*(y)\rangle_D = x^{\mathsf T}(DA^*)y$ , the adjoint’s matrix is

A^* = D^{-1}A^{\mathsf T}D = \begin{pmatrix} 1 & 0\\ 0 & \tfrac12\end{pmatrix} \begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & 2\end{pmatrix} = \begin{pmatrix} 0 & 0\\ \tfrac12 & 0\end{pmatrix} \neq A^{\mathsf T} .

Sanity check on $x = (1,0)$ , $y = (0,1)$ :

\langle u(x), y\rangle_D = \langle (0,0), y\rangle_D = 0, \quad \langle x, u^*(y)\rangle_D = \langle(1,0), (0,\tfrac12)\rangle_D = 0 ;

on $x = (0,1)$ , $y = (1,0)$ :

\langle u(x), y\rangle_D = \langle(1,0),(1,0)\rangle_D = 1, \quad \langle x, u^*(y)\rangle_D = \langle(0,1),(0,\tfrac12)\rangle_D = 1 .

Closing insight: “ $\operatorname{Mat}(u^*) = \operatorname{Mat}(u)^{\mathsf T}$ ” is a statement about orthonormal bases only; in general the metric $D$ intervenes, exactly as in the simultaneous reduction of the weekend problem.

Read in context →