# The Unapologetic Mathematician

## Mathematics for the interested outsider

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.