Positive definite matrix property

Question

Let $\lambda \in (0,1)$ and let $Q$ be an orthogonal matrix. Define

$a^*=Q^T\operatorname{diag} \{ \lambda/2,...,\lambda/2,2\lambda^{-1},...2\lambda^{-1}\} Q$.

Show that

$\frac{1}{2}\lambda I \leq a^* \leq 2\lambda^{-1}I$

Answer 1

Let me handlet one side for you and you should try it for another side

\begin{align}a^*-\frac12\lambda I &= Q^Tdiag(\frac{\lambda}{2}, \ldots,\frac{\lambda}{2}, \frac2\lambda , \ldots, \frac2\lambda )Q -\frac{\lambda}{2}I\\ &=Q^Tdiag(\frac{\lambda}{2}, \ldots,\frac{\lambda}{2}, \frac2\lambda , \ldots, \frac2\lambda )Q -Q^Tdiag(\frac{\lambda}{2}, \ldots,\frac{\lambda}{2}, \frac\lambda2 , \ldots, \frac\lambda2 )Q\\ &=Q^Tdiag(0, \ldots, 0, \frac2\lambda - \frac\lambda2, \ldots, \frac2\lambda-\frac\lambda2)Q\end{align}

Now, check that $\frac2\lambda - \frac\lambda 2 > 0$.

You might like to let $y=Qx$ and we have

$$x^T(a^*-\frac\lambda 2 I)x=y^Tdiag(0, \ldots, 0, \frac2\lambda - \frac\lambda 2, \ldots, \frac2\lambda-\frac\lambda2)y \ge 0$$