<s>[INST] Find $T^*\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ in an inner product space $V=\mathbb{R^3}$
$f$ is an inner product ,$f\Bigg (\begin{pmatrix}
           x_{1} \\
           x_{2} \\
           x_{3}
         \end{pmatrix} ,\begin{pmatrix}
           y_{1} \\
           y_{2} \\
           y_{3}
         \end{pmatrix} \Bigg ) = 4x_1y_1+x_1y_2+x_2y_1+4x_2y_2+x_2y_3+x_3y_2+x_3y_3$
$T\begin{pmatrix}
           a \\
           b \\
           c
         \end{pmatrix} =\begin{pmatrix}
           b+c \\
           a+b \\
           a+2c
         \end{pmatrix}$
Find $T^*\begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}$
My solution:
I want to find $T^*\begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}$ without finiding orthonormal basis. ( Is it possible ? )
Suppose $T^*\begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}= \begin{pmatrix}
           x \\
           y \\
           z
         \end{pmatrix}$
$\langle T\begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix} , \begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}\rangle$
$=\langle \begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}, T^*\begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix} \rangle $
$\langle T\begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix} , \begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}\rangle$ =$\langle \begin{pmatrix}
           5 \\
           3 \\
           7
         \end{pmatrix} , \begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}\rangle = 141$
Then, $\langle \begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}, T^*\begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix} \rangle = \langle \begin{pmatrix}
           1 \\
           2 \\
           3
         \end{pmatrix}, \begin{pmatrix}
           x \\
           y \\
           z
         \end{pmatrix} \rangle = 4x+y+2x+8y+2z+3y+3z=141$
Here I get stuck, how can I find $x,y,z$?
Am I supposed to take $x=y=0$ and then get Z and that's it?
Please any help or suggestions. Thanks.
 [/INST] If it is the word "orthonormal" that is bugging you then you can do it by taking a basis, say $v_{1},v_{2},v_{3}$. Where you can pick them to be any basis you like. Say $(0,1,1),(1,0,1),(1,1,0)$. Or heck, you might even take the standard basis for $\mathbb{R}^{3}$ , $e_{1},e_{2},e_{3}$. They are not orthonormal wrt the inner product given.
Then if $T^*(1,2,3)=(x,y,z)$. You have to solve the following system of equations for $(x,y,z)$.
$$\langle (1,2,3),T(v_{i})\rangle = \langle (x,y,z),v_{i}\rangle\,\, ,1\leq i\leq 3.$$ (As I said, if you want to make your work easier it suffices to work with $e_{1},e_{2},e_{3}$.)
They will give you three equations. You solve them to find $(x,y,z)$. And hence you will have your image.
</s>