Inner product question We are given an inner product of $\mathbb R^3$:
$f\left(\begin{pmatrix} x_1\\x_2\\x_3\end{pmatrix},\begin{pmatrix} y_1\\y_2\\y_3\end{pmatrix}\right) = 2x_1y_1+x_1y_2+y_2x_1+2x_2y_2+x_2y_3+y_3x_2+x_3y_3$
We are given a linear transformation $T$ such that:
$$T\begin{pmatrix} \;\;1\\\;\;0\\-1\end{pmatrix}=a\begin{pmatrix} \;\;1\\\;\;0\\-1\end{pmatrix}$$
$$T\begin{pmatrix}\;\;0\\\;\;1\\-1\end{pmatrix}=b\begin{pmatrix}\;\;0\\\;\;1\\-1\end{pmatrix}$$
$$T\begin{pmatrix}\;\;1\\-1\\\;\;2\end{pmatrix}=c\begin{pmatrix}\;\;1\\-1\\\;\;2\end{pmatrix}$$
Show that $T=T^*$ with respect to $f$ if and only if $a=b$. Meaning, for all $v\in \mathbb R^3$, $f(Tv,v)=f(v,Tv)$ if and only if $a=b$
The intuition says let's do Gram-Schmidt, look at the matrix of $T$ with respect to that orthonormal basis, transpose it, and see they are equal if $a=b$. In practice, that didn't work.

Note that the vectors you're given the effect of T on are a basis for $ \mathbb{R}^3$ of eigenvectors of T. Since everything is linear, you need only prove that $f(Tv,w) = f(v,Tw)$ iff $a = b$ for $v, w$ a pair (that could be equal) of these basis vectors. Note that you do need the equation to hold with the vectors used on each side not equal.
Try substituting some of these combinations in and see if you get anywhere. I think (but haven't actually written out the algerba) that it should work.
