Multiplying $A\preceq B$ with a matrix

Question

I have a matrix inequality, $$A\preceq B,$$ where $\preceq$ means that $B-A$ is psd.

update: How can I show that if $M$ is a positive definite matrix, then the inequality above is equivalent
$$M A M^\ast \preceq M B M^\ast.$$

Answer 1

Your "only if " is not ok here, for counter example take A=I and B=2I, I is Identity matrix here.
Then M can be any orthogonal matrix!
For "if" part ( what have you asked in comment?),.
Since, $M$ is positive definite iff $x^\ast Mx>0$ for all $x\neq0$. Write
$$xM^\ast(A-B)Mx^\ast=(Mx^\ast)^\ast (A-B) (Mx^\ast),$$ and denote $Mx^\ast=y$, and $y\neq 0$ because if $y=0$, then $xy=xMx*=0$. Now $y^\ast(A-B)y>0$ for all $y>0$, by definition of $(A-B)$ being positive definite. So, $M*(A-B)M$ positive definite if $M$ is positive definite.

Answer 2

If $M$ is any matrix of compatible dimensions, you certainly have $MAM^* \preceq MBM^*$. This is because: $$ x^*(MAM^*)x = (M^*x)^*A(M^*x) \leq (M^*x)^*B(M^*x) = x^*(MBM^*)x, $$ for any vector $x$. (The inequality above is immediate from $A \preceq B$.)