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Assume $T$ is an operator on $\mathbb R^3$, $B=\{(1, 1, 1)^T, (1, -1, 0)^T, (0, 1, -1)^T\}$ is a basis of eigenvectors for $T$ and that the corresponding eigenvalues of $T$ are the real numbers $a$, $b$, $c$. Prove that $T$ is self adjoint if and only if $b=c$. Thanks!

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As a complement to copper.hat's answer, which is what you should actually do, here's an excessively high tech, shamelessly spectral-theoretic approach that might nonetheless be instructive.

First, suppose that $T$ is self-adjoint. Then eigenvectors corresponding to different eigenvalues must be orthogonal to each other. Since $(1,-1,0)^T$, corresponding to $b$, and $(0,1,-1)^T$, corresponding to $c$, are evidently not orthogonal to each other, it therefore follows that they must correspond to the same eigenvalue, i.e., $b=c$.

Now suppose that $b=c$. Observe that $(1,1,1)^T$ is orthogonal to $(1,-1,0)^T$ and $(0,1,-1)^T$, so that if $E := \operatorname{span}\{(1,-1,0)^T,(0,1,-1)^T\} = \ker(T-bI)$, then $E^\perp = \operatorname{span}\{(1,1,1)^T\} = \ker(T-aI)$, and hence $T = aP_{E^\perp} + bP_E$, where $P_E$ denotes the orthogonal complement onto $E$ and $P_{E^\perp} = I - P_E$ denotes the orthogonal complement onto $E^\perp$. Since $P_E$ and $P_{E^\perp}$ are self-adjoint and $a$ and $b$ are real, it therefore follows that $T$ is self-adjoint.

Again, this is probably not the way you actually want to solve the problem, particularly showing that if $b=c$ then $T$ is self-adjoint, but you might find this perspective worth thinking about.

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    Oh, this is entirely how I'd think about it. The question is just how much technology the OP has at their disposal. Indeed, it's an *extremely* sophisticated undergraduate linear algebra course, at least by North American standards, that would formulate the finite-dimensional spectral theorem in terms of the orthogonal projections onto the eigenspaces. I was lucky enough to have this as a first year/second year undergrad, in a two-semester sequence using Friedberg/Insel/Spence, but I've since learnt this is not so common at all...2013-03-21
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Let $T_k = B E_{kk} B^{-1}$, where $E_{kk} = e_k e_k^T$, for $k=1,2,3$.

Then $T = a T_1 + b T_2 + c T_3$, and $T=T^T$ iff $a T_1 + b T_2 + c T_3 = a T_1^T + b T_2^T + c T_3^T$.

If you crank through the tedious calculations, you find

$T_1 = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ $T_2 = \frac{1}{3}\begin{bmatrix} 2 &-1 &-1 \\ -2 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ $T_3 = \frac{1}{3}\begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 &-2 \\ -1 &-1 & 2 \end{bmatrix}$

First we notice that $T_2=T_3^T$, and $T_1=T_1^T$. Hence if $b=c$, then $T=T^T$. Now suppose that $T=T^T$. Then we have $3 T_{31} = a-c = 3 T_{13} = a-b$, from which it follows that $b=c$.