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Let $A$ be a real symmetric matrix and $\lambda$ be its eigenvalue.

For a unit vector $x$, if $x^TAx=\lambda$, is it true that $Ax=\lambda x$?

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up vote 5 down vote accepted

No. Let $A$ be the diagonal matrix with entries $2,1,0$. Let $X = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 1/\sqrt{2}\end{pmatrix}$. Clearly $X$ is a unit vector.

$1$ is an eigenvalue of $A$. $X^TAX = 1$. But $AX = \begin{pmatrix} \sqrt{2} \\ 0 \\ 0\end{pmatrix}$.

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