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Let $x \in \Bbb R^n$ and $Q \in M_{n \times n}(\Bbb R)$, where $Q$ is hermitian and negative definite. Let $(\cdot,\cdot)$ be the usual euclidian inner product.

I need to prove the following inequality:

$$(x,Qx) \le a(x,x),$$

where $a$ is the maximum eigenvalue of $Q$.

Any idea?

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$Q$ can be unitarily diagonalized since it is Hermitian. What does the left hand side look like if $Q$ is diagonal? – copper.hat Jul 17 '12 at 18:16
up vote 1 down vote accepted

So, since $Q$ is Hermitian, it can be written as $Q=U\Lambda U^\text{H}$, where $\Lambda$ contains the eigenvalues (which are negative) and U is unitary ($U^HU=I$). Then

\begin{equation} (x,Qx) = x^H U\Lambda U^H x = (U^Hx) \Lambda (U^Hx) \end{equation}

We are done actually: since all eigenvalues are negative we have

\begin{equation} (U^Hx) \Lambda (U^Hx) \leq (U^Hx) A (U^Hx) \end{equation} where $A=\text{diag}(a,a,\ldots,a)=aI$. So

\begin{equation} (x,Qx) \leq a (U^Hx) (U^Hx) = a ||U^Hx|| = a||x||=a(x,x) \end{equation}

Here I also used the fact that, since U is unitary it does not affect the length of vectors.

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