$Q \leq_{S^{n}} 0$ if and only if $Q \leq_{S^{n}} sI$ for some $s < 0$

Question

Let $A,B$ be any square $n \times n$ matrices. Define $A \leq_{S^{n}} B$ if $A-B$ is negative semi-definite.

Given a matrix $Q$, is it true that $Q \leq_{S^{n}} 0$ if and only if $Q \leq_{S^{n}} sI$ for some $s < 0$ where $I$ is identity matrix?

I don't mind having a proof but I'm really looking for whether this is true or not, and if it isn't, then is there a way to flip the signs etc. to make it true?

Answer 1

No. Consider $Q=\pmatrix{0\\ &-1}$.