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Let $v_1$ and $v_2$, with $v_1\neq v_2$ and $\langle v_1,v_2\rangle\neq 0$, be two non zero vectors in an $N$ dimensional Hilbert space. Let $\mathbf{S}=v_1\otimes v_1 +v_2\otimes v_2$ be an $N\times N$ Hermitian matrix ($\otimes$ denoting outer product).

$\mathbf{S}$ has at most 2 non-zero eigenvalues. Assuming there are 2 distinct eigenvalues and denoting the two associated eigenvectors as $e_1$ and $e_2$, can it be said with certainty that $\langle e_1, v_1\rangle\neq 0$ and $\langle e_1,v_2\rangle\neq 0$ (and similarly for $e_2$)?

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This is equivalent to asking that $v_1, v_2$ are not the eigenvectors, and this follows from the conditions you've given. – Qiaochu Yuan Aug 19 '12 at 5:44
The eigenvalue and eigenvector will change in different basis, so you should point out the basis. Actually, they differ by something like $A=P^T\tilde{A}P$, rather $A=P^{-1}\tilde{A}P$. So if $P\not\in O(n)$, the eigenvalue and eigenvector may change. – user18537 Aug 19 '12 at 8:09
up vote 2 down vote accepted

By choosing a basis, one may assume wlog that $v_1 = (a,0,\ldots,0)^\top$ and $v_2 = (b,c,0,\ldots,0)^\top$ where $a = \|v_1\|$, $b = \langle v_1,v_2 \rangle/a$, $c = \sqrt{\|v_2\|^2 - b^2}$. Note that $a,b,c$ are all nonzero. Then $S$ has its top left $2 \times 2$ submatrix $$\pmatrix{a^2 + b^2 & bc \cr bc & c^2\cr}$$, the other entries being all $0$. We may as well ignore the rest of the matrix and assume $N=2$. The nonzero eigenvalues are the roots of the characteristic polynomial $\lambda^2 - (a^2 + b^2 + c^2) \lambda + a^2 c^2$. They are indeed distinct since $(a^2 + b^2 + c^2)^2 - 4 a^2 c^2 = (a^2 - c^2)^2 + 2 b^2 (a^2 + c^2) + b^4 > 0$. The eigenvector for eigenvalue $r$ is $x = (x_1,x_2)^\top$ where $(a^2 + b^2 - r) x_1 + b c x_2 = 0$. Now $\langle x, v_1 \rangle = a x_1 \ne 0$. By symmetry (we could have presented $v_1$ before $v_2$), $\langle x, v_2 \rangle \ne 0$ as well.

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