question about inner product and $f^*$

Question

In $\mathbb{R}$3 we declare an inner product as follows: $\langle v,u \rangle \:=\:v^t\begin{pmatrix}1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 3\end{pmatrix}u$

we have operator $f \colon V \to V$ , $f\begin{pmatrix}x \\y \\z\end{pmatrix}\:=\begin{pmatrix}1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9\end{pmatrix}\begin{pmatrix}x \\y \\z\end{pmatrix}$

The question is : calculate $f^*$.

So far, as i know, i need to find orthonormal basis $B$, and find $\left[f\right]_B^B$, and after that just do transpose to $\left[f\right]_B^B$.
is That correct? it's a question from test that i had and i didn't know how to answer it so i forwarding this to you. tnx!

Answer 1

Let us suppose that $V$ is a vector space over the reals. If $f(u)=Bu$ and $\langle v, u\rangle:= v^t Au$, with $B$ and $A$ $3\times 3$ matrices defined in the OP, then we want to find the matrix $Q$ s.t. $f^*(u):=Qu$ and

$$\langle v, f(u)\rangle\stackrel{!}{=} \langle f^*(v), u\rangle,$$

for all $u,v \in V$, i.e.

$$ v^t ABu\stackrel{!}{=} (Qv)^t Au = v^t Q^tA u, $$

for all $u,v,\in V$. As $A$ is invertible, then

$$AB=Q^tA \Leftrightarrow ABA^{-1}=Q^t,$$

or $Q=(ABA^{-1})^t$.

Answer 2

When A and B are two matrix and there exists an nonsingular matrix like P that $P^{-1}$AP=B then they are representation of one fixed operator but maybe respect to diffrent basis. so when you calculate $\left[f\right]_B^B$ by the orthonormal basis $B$ and transpose it , you have found one of the representations of $f^*$ . now for finding the $f^*:V\rightarrow\ V$ explicitly must do it in the basis that $f:V\rightarrow\ V$ has been done.

The easiest way to find $f^*:V\rightarrow\ V$ is , by the help of definition of it. the formula for the $f^*:V\rightarrow\ V$ is $$<f(x),y>=<x,f^*(y)>$$ now it is sufficient to find values of $f^*$ one the basis and extend it linearly to find the explicit formula for $f^*$