1

I want to find gradient of $\ln(\|Ax\|_2^2)$ where $A \in \mathbb{R}^{n \times n} $ and $x \in \mathbb{R}^{n}$.

$\ln(\|Ax\|_2^2) = \ln((Ax)^{T}Ax) = \ln(x^TA^TAx)$

And I am stuck here. Trying to do it like in one dimensional case would lead to something like $$\frac{1}{x^TA^TAx} * (x^T(A^TA + A^TA))$$

Is it right?

Kate
  • 191
  • That is correct, although many would say it's a column vector instead of a row vector. – greg Aug 04 '20 at 16:48
  • Perhaps use the number $2$ rather than repeating $A^T A$ :-). As @greg mentioned, the above is usually called the derivative, the gradient is a element of the ambient (inner product) space. Here that just means the transpose. – copper.hat Aug 04 '20 at 16:56

1 Answers1

0

Your answer is correct, and it follows from the chain rule. I set the problem up as follows to get the solution:

Let $\phi(\mathbf{x})=A\mathbf{x}$, $\psi(\mathbf{y})=\|y\|^2_2$ and $f(z)=\log(z)$. The expression in the problem is

$g(\mathbf{x})=(f\circ \psi\circ \phi)(\mathbf{x})$

$$ f'(z)=\frac{1}{z}$$ $$\psi'(\mathbf{y})=2\mathbf{y}^\top$$ $$\phi'(\mathbf{x})=A$$

Putting things together under the chain rule

$$ \begin{align} g'(\mathbf{x})&=f'(\psi(\phi(\mathbf{x}))\,\psi'(\phi(\mathbf{x}))\,\phi'(\mathbf{x})\\ &=\frac{1}{\|A\mathbf{x}\|^2_2}2(A\mathbf{x})^\top A\\ &=\frac{2}{\mathbf{x}^\top A^\top A\mathbf{x}}{\mathbf{x}^\top}A^\top A \end{align} $$

Mittens
  • 39,145