Let $A$ be a positive definite $n$ by $n$ matrix. Let $c_1$ be the smallest eigenvalue of $A$ and $c_2$ be the largest eigenvalue of $A$. Then how can I show that

$$ | \langle z, Aw \rangle |^2 \le \langle z, Az \rangle \langle w, Aw \rangle$$ and $$ c_1 |z|^2 \leq \langle z, Az \rangle \leq c_2 |z|^2$$ for any $z, w \in \mathbb C^n$?

Here $\langle x, y \rangle = \sum_{j=1}^n x_j \bar y_j$.


As $A$ is positive definite so $\exists P$ an unitary matrix such that $P^*AP=D=$diag $(\lambda_1,\lambda_2,\dots,\lambda_n)$,$\lambda_i$ are the eigen values of $A$ and $\lambda_i>0$

Let $P^*z=P^{-1}z=y=\begin{pmatrix} y_1&y_2&y_3&\dots &y_n\\ \end{pmatrix}^t,P^*w=P^{-1}w=x=\begin{pmatrix} x_1&x_2&x_3&\dots &x_n\\ \end{pmatrix}^t$(As $PP^*=I\Rightarrow P^{-1}=P^*$)

Then we have,


It is easy to check that $y^*Dx=\sum_{i=1}^{n}\bar{y_i}\lambda_ix_i$

Now by applying Cauchy Schwarz to the above sum we have,

$\displaystyle (\sum_{i=1}^{n}\bar{y_i}\lambda_ix_i)^2\le (\sum_{i=1}^{n}(\bar{y_i}\sqrt{\lambda_i})^2)(\sum_{i=1}^{n}({x_i}\sqrt{\lambda_i})^2)\le(\sum_{i=1}^{n}(\bar{y_i}\lambda_iy_i)(\sum_{i=1}^{n}(\bar{x_i}\lambda_ix_i)=<y,Dy><x,Dx>=<z,Az><w,Aw>$(As $PP^*=I$)

And the 2nd one,

$<w,Aw>=\sum_{i=1}^{n}\bar{x_i}\lambda_ix_i=\sum_{i=1}^{n}\lambda_i|x_i|^2\le c_2\sum_{i=1}^{n}|x_i|^2=c_2<x,x>=c_2<P^*w,P^*w>=c_2<w,w>=c_2\sum_{i=1}^{n}|w_i|^2$(As $PP^*=I$)

Similarly we also have,

$<w,Aw>=\sum_{i=1}^{n}\bar{x_i}\lambda_ix_i=\sum_{i=1}^{n}\lambda_i|x_i|^2\ge c_1\sum_{i=1}^{n}|x_i|^2=c_1<x,x>=c_1<P^*w,P^*w>=c_1<w,w>=c_1\sum_{i=1}^{n}|w_i|^2$(As $PP^*=I$)


Hint: As $A$ is positive, there is an orthogonal matrix $Q$ such that $A = Q^*\Lambda Q$ with $\Lambda = {\rm diag}(\lambda_1, \ldots, \lambda_n)$ the eigenvalues. Now for $w, z \in \mathbb C^n$, we have $$ \langle w, Az\rangle = \langle Qw, \Lambda Qz\rangle = \sum_{i} \lambda_i (Qw)_i\overline{(Qz)_i} $$ Now apply Cauchy-Schwarz for the first part, and $c_1 \le \lambda_i \le c_2$ for the second.


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