Mathematical Reflections 2014, Issue 5 - Problem U313

Problem:
Let $X$ and Y be nonnegative definite Hermitian matrices such that $X-Y$ is also non-negative definite. Prove that $\textrm{tr}(X^2) \geq \textrm{tr}(Y^2)$.

Proposed by Radouan Boukharfane

Solution:
We use Klein's Inequality:
For all nonnegative-definite Hermitian $n \times n$ matrices $A$ and $B$, and all differentiable convex functions $f:\mathbb{R}^+ \longrightarrow \mathbb{R}$, the following inequality holds:
$${\rm tr}[f(A)- f(B)- (A - B)f'(B)] \geq 0.$$

Take $A=X, B=Y$ and $f(x)=x^2$. By Klein's Inequality, we get $$\textrm{tr}\left(X^2-Y^2-(X-Y)\cdot2Y\right) \geq 0.$$ Moreover, since $X$ and $X-Y$ are nonnegative, then $$\textrm{tr}\left((X-Y)Y\right) \geq 0.$$ This can be proved by Cholesky decomposition: $X-Y=L_1L_1^*$ and $Y=L_2L_2^*$ for some lower triangular matrices $L_1,L_2$ with real and nonnegative diagonal entries. So, $$\textrm{tr}\left((X-Y)Y\right)=\textrm{tr}(L_1L_1^*L_2L_2^*)=\textrm{tr}(L_2^*L_1L_1^*L_2)=\textrm{tr}((L_1^*L_2)^*(L_1^*L_2))\geq 0.$$ Hence,
$$\begin{array}{lll}\textrm{tr}\left(X^2\right)-\textrm{tr}\left(Y^2\right)&=&\textrm{tr}\left(X^2-Y^2\right)\\&=&\textrm{tr}\left(X^2-Y^2-(X-Y)\cdot2Y\right)+2\textrm{tr}\left((X-Y)Y\right) \geq 0, \end{array}$$ which is the desired conclusion.