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?

  • $\begingroup$ 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 $Fx \neq 0$? $\endgroup$
    – Ralff
    Sep 29 '20 at 6:28
  • $\begingroup$ @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 $m$. $\endgroup$
    – Arthur
    Sep 29 '20 at 6:33
  • $\begingroup$ Oh yes. That is right. Thanks! This is obvious now. $\endgroup$
    – Ralff
    Sep 29 '20 at 6:34

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