Properties of Adjoints

Many of the properties of the adjoint construction follow immediately from the contravariant functoriality of the duality we used in its construction. But they can also be determined from the adjoint relation

$\displaystyle\left\langle T(v),w\right\rangle_W=\left\langle v,T^*(w)\right\rangle_V$

For example, if we have transformations $S:U\rightarrow V$ and $T:V\rightarrow W$ , then the adjoint of their composite is the composite of their adjoints in the opposite order: $(TS)^*=S^*T^*$ . To check this, we write

$\displaystyle\begin{aligned}\left\langle\left[TS\right](u),w\right\rangle_W&=\left\langle T\left(S(u)\right),w\right\rangle_W\\&=\left\langle S(u),T^*(w)\right\rangle_V\\&=\left\langle u,S^*\left(T^*(w)\right)\right\rangle_U\\&=\left\langle u,\left[S^*T^*\right](w)\right\rangle_W\end{aligned}$

It’s pretty straightforward to see that $I_V^*=I_V$ . Then, since $T^{-1}T=I_V$ and $TT^{-1}=I_W$ , we find that $T^*(T^{-1})^*=I_V$ and $(T^{-1})^*T^*=I_W$ , which shows that $(T^{-1})^*=(T^*)^{-1}$ .

Similarly, it’s easy to show that $(A+B)^*=A^*+B^*$ . But the process isn’t quite linear. When we work over the complex numbers, we find that $(\lambda T)^*=\bar{\lambda}T^*$ :

$\displaystyle\begin{aligned}\left\langle\left[\lambda T\right](v),w\right\rangle_W&=\left\langle T(\lambda v),w\right\rangle_W\\&=\left\langle\lambda v,T^*(w)\right\rangle_V\\&=\left\langle v,\bar{\lambda}T^*(w)\right\rangle_V\\&=\left\langle v,\left[\bar{\lambda}T^*\right](w)\right\rangle_V\end{aligned}$

Now if we restrict our focus to endomorphisms of a single vector space $V$ , we see that the adjoint construction gives us an involutory (since $T^{**}=T$ ), semilinear (since it applies the complex conjugate to scalar multiples) antiautomorphism of the algebra of endomorphisms of $V$ . That is, it’s like an automorphism, except it reverses the order of multiplication.

In a way, then, the adjoint behaves sort of like the complex conjugate itself does for the algebra of complex numbers (over the complex numbers we don’t notice the order of multiplication, but work with me here, people). This analogy goes pretty far, as we’ll see.

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May 25, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

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[...] linear. That is, we know that the matrix element can also be written as . And we also have the adjoint relation . Putting these together, we [...]

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[...] Self-Adjoint Transformations Let’s now consider a single inner-product space and a linear transformation . Its adjoint is another linear transformation . This opens up the possibility that might be the same transformation as . If this happens, we say that is “self-adjoint”. It then satisfies the adjoint relation [...]

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