Norm question (Frobenius norm)

[math8]

We show that if P and Q are Hermitian positive definite matrices satisfying

$$x^{*}Px \leq x^{*}Qx$$ for all $$x \in \textbf{C}^{n}$$

then $$\left\| P \right\|_{F} \leq \left\| Q \right\|_{F}$$

where $$\left\| \cdot \right\|_{F}$$ denotes the Frobenius norm (or Hilbert-Schmidt norm)

If A is a mXn matrix, then the Frobenius norm of A is
$$\left\| A \right\|_{F} = \left( \sum ^{m}_{i=1} \sum ^{n}_{j=1} \left| a_{ij}\right| ^{2} \right) ^{1/2} = \left( \sum ^{n}_{j=1} \left\| a_{j}\right\| ^{2}_{2} \right) ^{1/2}$$

with $$a_{j}$$ being the j-th column of A.I can see that the matrix (Q-P) is positive semi-definite. But from there I am not sure how to proceed.

 

[lippyka]

Any help would be appreciated. Hint: Use the fact that for any Hermitian positive definite matrix A, we have \left\| A \right\|_{F}^{2} = trace(A^2)We start by noting that the matrix (Q-P) is Hermitian positive semi-definite. This implies that \begin{align*}x^{*}(Q-P)x \geq 0\end{align*}for all x in \textbf{C}^{n}.Now, using the fact that for any Hermitian positive definite matrix A, we have \left\| A \right\|_{F}^{2} = trace(A^2), and the assumption that x^{*}Px \leq x^{*}Qx, we can deduce that\begin{align*}trace((Q-P)^2) &= \left\| Q-P \right\|_{F}^{2} \\&= \sum_{i=1}^n \sum_{j=1}^n (Q_{ij} - P_{ij})^2 \\ &= \sum_{i=1}^n \sum_{j=1}^n \left[ x_i^*Qx_j - x_i^*Px_j \right]^2 \\&\leq \sum_{i=1}^n \sum_{j=1}^n \left[ x_i^*Qx_j + x_i^*Px_j \right]^2 \\&= \sum_{i=1}^n \sum_{j=1}^n \left[ x_i^*(Q+P)x_j \right]^2 \\&= \left\| Q + P \right\|_{F}^{2} \\&= trace((Q+P)^2)\end{align*}Therefore,\begin{align*}trace(P^2) &= trace((Q+P)^2) - trace((Q-P)^2) \\&\leq trace((Q+P)^2) \\&= \