# Can someone explain how this is not self adjoint?

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?

• There's a good chance that what you were told was actually: Trying with just two particular vectors is not enough to prove that the operator is self-adjoint. Nov 4, 2018 at 12:29

## 1 Answer

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.