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Let $A$ be a 3x3 real symmetric matrix, and let $v_1, v_2$ be orthogonal eigenvectors of $A$. Suppose $v_3$ is orthogonal to both $v_1$ and $v_2$. Is $v_3$ necessarily an eigenvector of $A$?

To put it another way: by the spectral theorem we know that $\mathbb{R}^3$ has an orthonormal basis consisting of eigenvectors of $A$, but can we extend any two orthogonal eigenvectors of $A$ into such a basis?

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up vote 4 down vote accepted

You don't need the spectral theorem to prove this. We are given that $\langle Av_1, v_3 \rangle = \langle Av_2, v_3 \rangle = 0$, and it follows directly by symmetry that $\langle v_1, Av_3 \rangle = \langle v_2, Av_3 \rangle = 0$. To make it really obvious, $A$ must be symmetric in the basis $\{ v_1, v_2, v_3 \}$, and in that basis we are given that it looks like

$$\left[ \begin{array}{ccc} \lambda_1 & 0 & * \\\ 0 & \lambda_2 & * \\\ 0 & 0 & * \end{array} \right]$$

and we can complete the rest of the matrix by symmetry.

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Yes: If we denote by $W\subset\mathbb{R}^3$ the span of $\{v_1,v_2\}$, then the possible choices of such $v_3$ are exactly the nonzero elements of the line $W^\perp$. Given that $A$ is real symmetric and $v_1,v_2$ are orthogonal eigenvectors, there must exist some such choice of $v_3$ which is an eigenvector, but then all other possible choices of $v_3$ are nonzero multiples of this one, hence also eigenvectors.

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