I have matrix that represent $T:V \to V$ (linear map over $\mathbb{R}$) according to basis $E$.
$E$ is not an orthonormal set.
how can I know if this $T$ is self-adjoint.
I know that if $E$ was orthonormal basis we would take the transpose matrix.
but what about here ?
the matrix is : $$\begin{bmatrix} 1 & 2 \\ 2 & 1 \\ \end{bmatrix}$$
On
@SandraWest, so we need to compute. (Or use the matrix short-cut in my comment.) Remember that $T(v_1)=v_1+2v_2$ and $T(v_2)=2v_1+v_2$. It suffices (by linearity) to check whether $\langle Tx,y\rangle=\langle x,Ty\rangle$ when $x$ and $y$ are basis vectors. You can see the only thing that really requires checking is whether $\langle Tv_1,v_2\rangle = \langle v_1,Tv_2\rangle$.
Well, \begin{align*} \langle Tv_1,v_2\rangle & =\langle v_1+2v_2,v_2\rangle=−1+2=1, \quad\text{and} \\ \langle v_1,Tv_2\rangle &= \langle v_1,2v_1+v_2\rangle = 4−1=3 \,, \end{align*}
so T is not self-adjoint.
Without an orthonormal matrix I'm afraid you'll have to do it the long way: we're looking for a matrix $\,T^*\,$ s.t. (I'm assuming the usual, euclidean inner product):
$$\overline x:=\binom xy\,,\,\,\overline y=\binom ab\implies\langle T\, \overline x\,,\,\overline y\rangle=\langle \overline x\,,\,T^*\,\overline y\rangle\;\iff $$
$$\left\langle\,\begin{pmatrix}1&2\\2&1\end{pmatrix}\binom xy\,,\,\binom ab\,\right\rangle =\left\langle\,\binom xy\,,\,\begin{pmatrix}p&q\\r&s\end{pmatrix}\binom ab\,\right\rangle\iff$$
$$\left\langle\,\binom{x+2y}{2x+y}\,,\,\binom ab\,\right\rangle=\left\langle\,\binom xy\,,\,\binom{pa+qb}{ra+sb}\,\right\rangle\iff$$
$$ax+2ay+2bx+by=pax+qbx+ary+bsy\iff$$
$$(1-p)ax+(1-s)by+(2-r)ay+(2-q)bx=0$$
As the above has to be true for any two vectors $\,\overline x\,,\,\overline y\,$ , you can now choose particular cases and deduce the possible values of $\,p,q,r,s\,$ (which, of course, you then need to check in the general case), for example: with $\,x=a=1\,,\,y=b=0\,,\,$ :
$$(1-p)=0\implies p =1\;\;,\;\;etc.\ldots$$