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How can I prove the following?

If $A$ is $n \times n$ hermitian positive definite matrix, then there exists $c >0$ such that $$ (Ax) \cdot \bar x \leqslant c |x|^2 $$ where $x$ is an $n$-dimensional complex valued column matrix.

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What does Cauchy-Schwartz give? –  TenaliRaman Jun 3 '12 at 20:23
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2 Answers

up vote 1 down vote accepted

If a matrix is hermitian, then all its eigen values are real. Further since $A$ is given to be positive definite, all the eigenvalues are positive. Order the eigenvalues as $$\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$$ Also, since the matrix $A$ is hermitian, the eigenvectors can be chosen to be orthonormal and hence will span $\mathbb{C}^n$. Call them $\vec{e_1},\vec{e_2}, \ldots ,\vec{e_n}$ where $\vec{e_k}$ corresponds to the $k^{th}$ eigenvalue.

Now since $\vec{e_k}$ are orthonormal and span $\mathbb{C}^n$, we have that any vector $x \in \mathbb{C}^n$ can be written as $$x = x_1 \vec{e_1} + x_2 \vec{e_2} + \cdots + x_n \vec{e_n}$$ Hence, $Ax = \lambda_1 x_1 \vec{e_1} + \lambda_2 x_2 \vec{e_2} + \cdots + \lambda_n x_n \vec{e_n}$. $$Ax \cdot x = \lambda_1 x_1^2 + \lambda_2 x_2^2 + \cdots + \lambda_n x_n^2$$ This is so since $\vec{e_k}$'s are orthonormal. Since $\lambda_n \leq \lambda_{n-1} \leq \cdots \leq \lambda_2 \leq \lambda_1$, we get that $$Ax \cdot x = \lambda_1 x_1^2 + \lambda_2 x_2^2 + \cdots + \lambda_n x_n^2 \leq \lambda_1 \left( x_1^2 + x_2^2 + \cdots + x_n^2\right) = \lambda_1 \lVert x\rVert_2^2$$

Hence, $c = \lambda_1$ does the job for us.

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Hint:

1) Consider continuous function $$ f:S^{n-1}\to\mathbb{R}:x\mapsto (Ax)\cdot\overline{x} $$ defined on a compact space $S^{n-1}$. It is achieves its maximum at some vector $x_0\in S^{n-1}$.

2) Use the following identity $$ \frac{(Ax)\cdot\overline{x}}{|x|^2}=f\left(\frac{x}{|x|}\right) $$

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