# Is left and right multiplying a positive semi-definite by a matrix result in a positive semi-definite matrix?

Consider an $$n \times n$$ positive semi-definite matrix $$\mathbf{P}$$. Consider a function $$f(\mathbf{x})$$ where $$\mathbf{x}$$ is a vector. Let the Jacobian of $$f$$ be defined as the $$n \times m$$ matrix $$\mathbf{F} = \frac{\partial f}{\partial \mathbf{x}}$$ where $$\mathbf{F}$$ is evaluated at some point $$\mathbf{x}$$.

If I left and right multiple $$\mathbf{P}$$ by $$\mathbf{F}$$, is the result positive semi-definite? In other words, is $$\mathbf{F}^T \mathbf{P} \mathbf{F}$$ positive semi-definite?

More generally, is this true of any $$n \times m$$ matrix $$\mathbf{F}$$?

Hint: Take a good look at $$x^TF^TPFx$$. Knowing that $$P$$ is positive semidefinite, what can you say?
• I see. So, by definition, we know $F^TPF$ is positive semidefinite as long as $Fx \neq 0$. So, I suppose we require that $F$ is full rank, so $Fx \neq 0$? Or, is something else required for $Fx \neq 0$? Sep 29 '20 at 6:28
• @Ralff Positive SEMI-definiteness allows the result to be 0. So no rank requirements or anything. Positive definiteness of $P$ is not preserved this way, unless we require that $F$ has rank $m$. Sep 29 '20 at 6:33