# Show that $v \cdot w = 0$ under specific conditions.

Let $S$ be a symmetrix $n \times n$ matrix. (Remember that means $S^t = S$). Let $v$, $w$ be eigenvectors of $S$ for eigenvalues $\lambda$, $\mu$ respectively. Suppose $\lambda \not= \mu$. Show that $v \cdot w = 0$.

Hints give:

• In other words, the eigenspaces of symmetric matrices are perpendicular to each other.
• Use a: Where you let $x, y \in \mathbb{R}^n$. Verify that $(Sx) \cdot y = x \cdot (Sy)$. Answer: I verified that they do equal.
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A better title would be the first item after "In other words". –  lhf Apr 3 '12 at 0:11

$\lambda (v\cdot w) = \lambda v \cdot w =Sv\cdot w = v\cdot Sw = v \cdot \mu w = \mu (v\cdot w)$ implies $(\lambda-\mu) (v\cdot w) = 0$.