Can someone explain how this is not self adjoint?

Question

The matrix representation of a linear operator T: $\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ is given with respect to the following basis \begin{bmatrix}-5 &2 & \\ 2&5 & \end{bmatrix} is the matrix of T when written with respect to the basis $(1,0)$ and $(1,1)$. We take the standard dot product as its inner product.

I'm getting that it is self-adjoint as $\langle Tv_1,v_2 \rangle = \langle v_1,Tv_2 \rangle$ where $v_1 = (1,0)$ and $v_2=(1,1)$ but I was told that it wasn't. Could someone please shed some light on this? Something about the conjugate transpose not being equal?

Answer 1

HINT: let $u:=u_1e_1+u_2e_2$ and $v:=v_1e_1+v_2e_2$ for some basis $e_1,e_2$ of $\Bbb R^2$ and $T$ a linear operator from $\Bbb R^2$ to itself, then from the linearity of $T$ and the definition of self-adjoint

$$\forall u,v\in\Bbb R^2:\langle Tu,v\rangle=\langle u,Tv\rangle \iff \forall j,k\in\{1,2\}:\langle Te_k,e_j\rangle=\langle e_k,Te_j\rangle$$

So you only need to check if the RHS of above holds or not holds for some basis of $\Bbb R^2$ and $T$ represented using the standard orthonormal basis of $\Bbb R^2$, because the standard inner product is defined using the standard orthonormal basis.