Can the following matrix $A=\begin{pmatrix} 0 & 1 \\ 2 & 1 \end{pmatrix}$ be a matrix of self-adjoint operator in inner product space in some (not necessarily orthonormal) basis?
My approach: It is easy to check that eigenvalues of this operator are $2,-1$. And we can easily find eigenspaces in each case, namely $V_{-1}=\langle (-1,1)\rangle$ and $V_{2}=\langle (1,2)\rangle $. Moreover, we know the following fact
Let $V$ be a finite-dimensional euclidean space and $f:V\to V$ is an operator. An operator $f$ is self-adjoint if and only if there is an orthonormal basis of eigenvectors.
Let's define in our space $V$ the following inner product: $$\langle x,y\rangle=(x_1,x_2)\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}, $$ where $x=(x_1,x_2), y=(y_1,y_2)$.
Let's denote $e_1=(-1,1)$ and $e_2=(1,2)$ then $e_1\perp e_2$ and $|e_1|=\sqrt{3}, |e_2|=\sqrt{6}$. Let's define new vectors $e'_1:=\dfrac{e_1}{\sqrt{3}}$ and $e'_2:=\dfrac{e_2}{\sqrt{6}}$. So we see that $\{e'_1,e'_2\}$ is orthonormal basis of eigenvectors. So it means that initial matrix can be matrix of self-adjoint operator.
Is my reasoning correct?
Would be very grateful for any comments!
