Prove T is symmetric if and only if b=c

Question

Assume T is a linear operator on R^3, that α={(1,1,1),(1,-1,0),(0,1,-1)} is a basis consisting of eigenvectors and that the corresponding eigenvalues of T are real numbers a,b,c. Prove that T is symmetric if and only if b=c.

I have proved that if T is symmetric, b=c; now I'm having difficulty proving from the reverse direction. If b=c, how should I start the prove?

I know that T is symmetric if ⟨T(x),y⟩=⟨x,T(y)⟩ for all x and y, but it seems not helping

Answer 1

Calculate the matrix elements of $T$ in the given basis. You should find that if $b = c$, the matrix is symmetric. That is, $\langle T\alpha_i, \alpha_j\rangle = \langle \alpha_i, T\alpha_j\rangle$ for $i, j = 1, 2, 3$.

From this you can show that $\langle T x, y\rangle = \langle x, T y\rangle$ for any vectors $x$ and $y$, by decomposing as $x = \sum_i{b_i\alpha_i}$ and $y = \sum_j{c_j\alpha_j}$.