I'm trying to find the derivative of $f(x)=\|Ax\|_2^2$ where $A$ is some matrix and $\|u\|_2$ is the euclidean norm of $u$, $\|u\|_2 = \sqrt{u_1^2+u_2^2+\cdots+u_n^2}$

I know how to do this by expanding the terms, using a lot of $\sum$ signs etc, but I am trying to avoid all that by using the definition.

$$f'(x) = \lim_{t \to 0} \frac{f(x+th)-f(x)}{t} = \lim_{t \to 0} \frac{\|Ax+Ath\|_2^2-\|Ax\|_2^2}{t}=\lim_{t \to 0} \frac{(\|Ax+Ath\|_2+\|Ax\|_2)(\|Ax+Ath\|_2-\|Ax\|_2)}{t}$$

And I'm stuck here, don't know how to proceed.

Is there a way of doing this without actually writing each norm as a sum and writing $Ax$ explicitly etc?

  • $\begingroup$ When you write a lim_{t\to0} b rather than a\lim_{t\to0} b in a "displayed" (as opposed to "inline") setting, then you see $\displaystyle a lim_{t\to0} b$ rather than $\displaystyle a\lim_{t\to0} b$. The three differences are (1) $\lim$ is not italicized; (2) proper spacing between $a$ and $\lim$ and between $\lim$ and $b$; and (3) the position of the subscript. That last one doesn't apply to inline settings. The latter form is standard and I changed it. $\endgroup$ – Michael Hardy May 13 '14 at 20:00

We have

$$f(x)=\langle Ax,Ax\rangle =\varphi\circ\psi(x)$$ where $$\varphi(u,v)=\langle u,v\rangle$$ is a bilinear map and $$\psi(x)=(Ax,Ax)$$ is a linear map, hence by the chain rule we have $$Df(x)h=D\varphi(\psi(x))D\psi(x)h=D\varphi(\psi(x))(Ah,Ah)=\langle Ax,Ah\rangle+\langle Ah,Ax\rangle\\=2\langle Ax,Ah\rangle$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.