Gradient of $f(x) = x^TP^{-1}x$

Question

I saw from this link how to calculate the gradient of the function f(x)=$x^TPx$:
How to calculate gradient of $x^TAx$

Does this proof still hold true when the inverse of the P-matrix is used? My assumption is that it should, as I could just let a matrix $A=P^{-1}$, and then follow the steps outlined in the link to arrive at $f'(x)=\nabla f=x^T(A+A^T)$ or for my specific question $f'(x)=\nabla f=x^T(P^{-1}+(P^{-1})^T)$

Are there any holes in this logic?

You are right, you can use the other result directly by substituting $A = P^{-1}$; nothing more is needed. — angryavian, Nov 07 '18 at 19:08

score 0 · Accepted Answer · answered Nov 07 '18 at 19:12

0

Does this proof still hold true when the inverse of the P-matrix is used?

Yes it does. Just replace $A = P^{-1}$ and you will get $$f'(x)=\nabla f=x^T(P^{-1}+P^{-T}) $$ where $P^{-T} = (P^{-1})^T = (P^T)^{-1}$.

answered Nov 07 '18 at 19:12

Ahmad Bazzi

12,076

Gradient of $f(x) = x^TP^{-1}x$

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