Let's say we have some linear transformation $T$.
If we express it in the form: $B^{-1} A B$, where $B$ maps us into some aritrary basis, and $B^{-1}$ maps us back out.
Is $A$ always symmetric? Or can it at least be converted to a symmetric matrix?
I got to the conclusion that $A$ is symmetric by using this argument: Why do we assume that a matrix in quadratic form is Symmetric?
And working through some small matrix examples, but looking to see if there is a simpler way to show this.
Over the reals this need not be possible.
Let $V:=V_2(\mathbb{R})$, and let $T:(1,0)\mapsto (0,1)$, $T:(0,1)\mapsto (0,0)$.
The characteristic polynomial is $X^2$.
Now look at an arbitrary symmetric matrix $\begin{pmatrix} a & b\\ b & c\end{pmatrix}$. Its characteristic polynomial is $X^2-(a+c)X+(ac-b^2)$, so it can only represent $T$ if $a=-c$ and $ac-b^2=0$: so $a=b=c=0$ which won't do.
Hint: See if that happens with $T \colon \Bbb{R}^2 \to \Bbb{R}^2$ such that $T(x,y) = (2x+4y,3x+5y)$ for all $(x,y)\in\Bbb{R}^2$ if you consider $\mathcal{B} =((1,0),(0,1))$ and $\mathcal{B}’ = ((1,1),(1,-1))$.
