Preservation of definiteness dominance

Question

Suppose that $A$ and $B$ are $n\times n$, symmetric, and positive definite. We write $A\succ 0$ and $B\succ 0$. Suppose that it is also true $A\preceq B$, that is, $B-A$ is positive semidefinite. I'm wondering if the following are true: $$ \forall D: n\times m:ADD'A\preceq BDD'B\tag{$*$} $$ and $$ \forall d: n\times 1:Add'A\preceq Bdd'B\tag{$**$} $$ My intuition is that we are essentially squaring $A$ and $B$, then multiplying the results by the same nonnegative entitiy. So the original relation between $A$ and $B$ themselves ($A\preceq B$) should be preserved and I expect ($*$) and ($**$) are both true but I cannot produce a proof.

Answer 1

Neither claim is true.

Let $m=n=2$.

Let $A,B$ be given by $$ A= \pmatrix{ 3&2\\ 2&3\\ },\;\;\; B= \pmatrix{ 5&1\\ 1&3\\ } $$ Then $A$ has characteristic polynomial $x^2-6x+5$, and $B$ has characteristic polynomial $x^2-10x+24$, so $A,B$ are both positive definite.

For $B-A$, we have $$ B-A= \pmatrix{ 2&-1\\ -1&2\\ } $$ which has characteristic polynomial $x^2-4x+3$, so is positive definite.

Letting $D$ be given by $$ D= \pmatrix{ 2&1\\ 1&1\\ } $$ we get $$ BDD'B-ADD'A= \pmatrix{ 68&32\\ 32&11\\ } $$ which has characteristic polynomial $x^2-79x-276$, so is not positive semidefinite.

Letting $d$ be given by $$ d= \pmatrix{ 1\\ 0\\ } $$ we get $$ Bdd'B-Add'A= \pmatrix{ 16&-1\\ -1&-3\\ } $$ which has characteristic polynomial $x^2-13x-49$, so is not positive semidefinite.

Thus, both claims are disproved.