Deducing self-adjointness from given $2\times 2$ matrix

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!