Show that a matrix A above the R field is congruent to itself squared iff A is non-negative

Question

How do I show that a symmetric matrix A above the R field is congruent to itself squared iff A is non-negative, but I'm not sure if by non negative the author meant positive semi-definite:

$$\exists P: P^tA^2P=A \iff \forall x: x^tAx\:\ge 0$$

or just non-negative:

$$\exists P: P^tA^2P=A \iff A \ge 0$$

i.e, for each element in the matrix it's positive.

Answer 1

It's always a good idea to start by figuring out what the question means.

Let's see: $$A = \pmatrix{1 & -1\cr -1 & 1\cr}$$ is positive semidefinite but not elementwise positive, and $$A^2 = \pmatrix{2 & -2\cr -2 & 2} = P^T A P$$ where $$ P = \pmatrix{\sqrt{2} & 0 \cr 0 & \sqrt{2}}$$ so it must be that positive semidefinite was meant.

Hint: Sylvester's law of inertia.