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Let $x$ and $y$ be unit vectors in the Euclidean norm. Define $s(\lambda) = |y^Hx|$, where $Ax=\lambda x$ and $ y^HA=\lambda y^H$. Here $\lambda$ is a simple eigenvalue (an eigenvalue with algebraic multiplicity $1$). I would like to prove $s(\lambda) \leq 1$ and $s(\lambda)\neq 0$. I tried the following and gave up after that.

$$Ax=\lambda x $$
Premultiply by $y^H$
$$\begin{align*} y^HAx &= \lambda y^Hx \\ y^HAx &= \lambda y^H I x \end{align*}$$
Taking norm on both sides,
$$ \begin{align*} \frac{1}{\lambda} \|y^HAx\| &= \|y^H I x \| \\ \frac{1}{\lambda} \|y^HAx\| &= |y^Hx| \end{align*}$$

Can anyone help me how to proceed after this?

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$s(\lambda)\le1$ is immediate from the fact that $x$ and $y$ are unit vectors; it has nothing to do with them being eigenvectors of $A$. So you only need to show $s(\lambda)\neq0$. –  joriki Nov 2 '11 at 7:52
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up vote 1 down vote accepted

In your attempt, you wrote $\frac1\lambda$, which is not a valid fraction if $\lambda=0$. Also, you haven't made use of the fact that $\lambda$ is a simple eigenvalue. Perhaps you should try another approach. For instance, assume that $x$ is a unit vector. If $U$ is a unitary matrix with $x$ as its first column (i.e. $U(1,0,\ldots,0)^\top = x$), then $$ A = U \begin{pmatrix} \lambda&v^\top\\ 0&B \end{pmatrix} U^\ast = U\widehat{A}U^\ast \textrm{ (say)} $$ for some vector $v$ and some matrix $B$. Since $U$ is unitary, eigenvalues, algebraic/geometric multiplicities of eigenvalues as well as orthogonality of vectors are preserved. Yet it is easier to make your argument by considering $\widehat{A}$. For example, that $\lambda$ is a simple eigenvalue means $B-\lambda I$ is invertible.

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