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I have

$f: R^n \Rightarrow R^1$

and $H$ a $n \times n$ matrix (that is inversible)

What's the gradient of:

$g: x \mapsto f(H.x)$

expressed using the gradient of f? Is it just $x \mapsto H. \nabla f(H.x) $ ?

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1 Answer 1

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You get $\nabla g(x) = H^T\cdot \nabla f(H\cdot x)$ from $\nabla g(x) = (D g(x))^T = (D f(Hx)\cdot H)^T = H^T \cdot (D f(Hx))^T = H^T \cdot \nabla f(Hx)$.

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  • $\begingroup$ Thanks, I knew I was doing something wrong... $\endgroup$
    – d--b
    Dec 1, 2016 at 8:47

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