$T: \mathbb R^n \to \mathbb R^n$, $\langle Tu,v\rangle=\langle u,T^*v\rangle$, is $T^*=T^t$ regardless of inner product?

Question

Basic question in linear algebra here. $T$ is a linear transform from $\mathbb R^n$ to $\mathbb R^n$ defined by $T(v)=Av$, $A\in \mathrm{Mat}_n(\mathbb R)$. We are given some inner product $\langle ,\rangle$ of $\mathbb R^n$. Does not have to be the standard one, just some random inner product.

let $T^*$ be a linear transform from $\mathbb R^n$ to $\mathbb R^n$ such that for all $u,v \in \mathbb R^n$: $\langle T(u),v\rangle=\langle u,T^*(v)\rangle$

I know that if $\langle ,\rangle$ is the standard inner product of $\mathbb R^n$, then $T^*(v)=A^tv$. My question is, does this hold for all inner products?

If it isn't, given some inner product and the transform $T$, how can I find $T^*$?

And if it is true, then why?

Answer 1

In general, the adjoint of $T$ is not given by multiplication with the transpose of the matrix of $T$.

You can represent an inner product $\langle\,\cdot\,,\,\cdot\,\rangle$ by a matrix, let's call it $S$,

$$\langle x,y\rangle = {}^tx Sy.$$

Then let $A$ be the matrix representing $T$, and $B$ the matrix representing $T^\ast$. By definition, you have, for all $x,y$,

$${}^tx({}^tA S) y = ({}^tx{}^tA) S y = {}^t(Ax) S y = \langle Tx,y\rangle = \langle x, T^\ast y\rangle = {}^t x S (B y) = {}^t x (SB) y,$$

and so you have

$${}^tAS = SB \iff B = S^{-1}{}^tAS.$$

Answer 2

No it does not hold for any inner product.

It is hard hard to show that any other inner product $\langle\cdot,\cdot\rangle_*$ can be represented as $$ \langle x,y\rangle_\star=\langle x,Sy\rangle, $$ where $S$ is a positive definite matrix. So $$ \langle Tx,y\rangle_\star=\langle x,T^*Sy\rangle \ne \langle x,ST^*y\rangle=\langle x,T^*y\rangle_\star, $$ unless $ST^*=T^*S$.