Problem: Let $T$ denote a linear map defined on $V=\mathbf{R}^{3}$, and let $(\cdot, \cdot)$ denote an inner product defined over $V$. Let $\mathcal{B}=\left\{v_{1}, v_{2}, v_{3}\right\}$ denote a basis for $V$. Assume that for all $1 \leq i \leq 3$ we have $\left(v_{i}, v_{i}\right)=1$, and that $\left(v_{1}, v_{2}\right)=1 / 2$ and $\left(v_{1}, v_{3}\right)=\left(v_{2}, v_{3}\right)=0$. Assume that $$ [T]_{\mathcal{B}}=\left(\begin{array}{lll} a & & \\ & b & \\ & & c \end{array}\right) $$ Prove that $T$ is self-adjoint if and only if $a=b$.
Attempt: From the matrix representation above we can deduce that $T v_{1}=a v_{1} ; T v_{2}=b v_{2}$ and $T v_{3}=c v_{3}$. Let $ v \in V $ be arbitrary. Write $ v= \alpha_{1} v_{1}+\alpha_{2} v_{2}+\alpha_{3} v_{3} $. Thus $ \left(\alpha_{1} T v_{1}+\alpha_{2} T v_{2}+\alpha_{3} T v_{3}, \alpha_{1} v_{1}+\alpha_{2} v_{2}+\alpha_{3} v_{3}\right)=\left(a \alpha_{1} v_{1}+b \alpha_{2} v_{2}+c \alpha_{3} v_{3}, \alpha_{1} v_{1}+\alpha_{2} v_{2}+\alpha_{3} v_{3}\right) $
Since $\left(v_{1}, v_{3}\right)=\left(v_{2}, v_{3}\right)=0$, then when we use the linearity of the inner product, 4 of the terms contribute zero. Thus we have $$ \begin{gathered} (T v, v)=a \alpha_{1} \alpha_{1}\left(v_{1}, v_{1}\right)+b \alpha_{2} \alpha_{2}\left(v_{2}, v_{2}\right)+c \alpha_{3} \alpha_{3}\left(v_{3}, v_{3}\right)+a \alpha_{1} \alpha_{2}\left(v_{1}, v_{2}\right)+b \alpha_{2} \alpha_{1}\left(v_{2}, v_{1}\right)= \\ =a \alpha_{1} \alpha_{1}+b \alpha_{2} \alpha_{2}+c \alpha_{3} \alpha_{3}+\frac{1}{2} a \alpha_{1} \alpha_{2}+\frac{1}{2} b \alpha_{2} \alpha_{1} \end{gathered} $$
Here we used the fact that $\left(v_{i}, v_{i}\right)=1$ and that $\left(v_{1}, v_{2}\right)=1 / 2$.
We can write $ (T v, v) = a \alpha_{1}^2+b \alpha_{2}^2 +c \alpha_{3}^2 +\frac{1}{2} a \alpha_{1} \alpha_{2}+\frac{1}{2} b \alpha_{2} \alpha_{1} $
[ From here I got stuck ]
I don't know how to continue or how to otherwise do the proof, I get stuck in both implications. Can you please help?
Assume $T$ is symmetric. Then $a,b$ are real eigenvalues. If $a\neq b$ the corresponding eigenvectors $v_1$ and $v_2$ are orthogonal because $$a\langle v_1,v_2\rangle =\langle Tv_1,v_2\rangle =\langle v_1,Tv_2\rangle =b\langle v_1,v_2\rangle$$ This gives a contradiction.