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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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+1: I like the use of not orthogonal to conclude $b=c$. – copper.hat Mar 21 '13 at 4:57
Right, nothing to be ashamed of, but not so high-tech either. Lots of people would do it like that, including me. +1. – 1015 Mar 21 '13 at 5:16
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... – Branimir Ćaćić Mar 21 '13 at 5:49

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 $b=c$.

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