Let $X,Y$ be real symmetric matrices. We say $X \geq Y$ if $X-Y$ is PSD.
Question: What is a basic ($2\times 2$) example of matrices $X,Y$ s.t.: $$ X \leq Y $$ but: $$ X^2 > Y^2 $$ e.g.$X^2 - Y^2$ is strictly positive definite? All I know is that $X,Y$ cannot commute but I am having difficulty coming up with concrete examples.
Here's an example if we allow arbitrary symmetric matrices: take $X = -2I$ and $Y = I$, where $I$ denotes the identity matrix. Of course, we have $X^2 = 4I \geq I = Y^2$.
No example exists in which $0 \leq X \leq Y$ but $X^2 \geq Y^2$. One way to see that this is the case is to note that $X \leq Y$ implies that $\lambda_i(X) \leq \lambda_i(Y)$ (where $\lambda_i(X)$ denotes the $i$th eigenvalue of $X$ in increasing order), and $X < Y$ implies $\lambda_i(X) < \lambda_i(Y)$. So, $X \leq Y$ implies that $\lambda_1(X) \leq \lambda_1(Y)$, which means that $$ \lambda_1(X^2) = \lambda_1(X)^2 \leq \lambda_1(Y)^2 = \lambda_1(Y^2). $$ However, $X^2 > Y^2$ necessitates that $\lambda_1(X^2) > \lambda_1(Y^2)$, which is contradicted in the above inequality.