question about inner product and $f^*$ 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!
 A: 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 $f^*$
A: 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 $Q=(ABA^{-1})^t$.
