# Gradient of l2 norm squared

Could someone please provide a proof for the following rule:

$$\nabla\|x\|_2^2 = 2x$$

I.E. why is the gradient of the $L_2$ norm square of $x$ equal to $2x$?

Thanks

Use the definition. If $$f(x)=\|x\|^2_2= \left(\left(\sum_{k=1}^n x_k^2 \right)^{1/2}\right)^{2}=\sum_{k=1}^n x_k^2 ,$$ then $$\frac{\partial}{\partial x_j}f(x) =\frac{\partial}{\partial x_j}\sum_{k=1}^n x_k^2=\sum_{k=1}^n \underbrace{\frac{\partial}{\partial x_j}x_k^2}_{\substack{=0, \ \text{ if } j \neq k,\\=2x_j, \ \text{ else }}}= 2x_j.$$ It follows that $$\nabla f(x) = 2x.$$

• Thanks. One follow up question, does this still hold if x is complex? – user167133 Jul 30 '14 at 20:56
• Isn't the L2 norm defined as: $$f(x)=\|x\|^2_2= \left(\left(\sum_{k=1}^n |x_k|^2 \right)^{1/2}\right)^{2}=\sum_{k=1}^n |x_k|^2$$, there is an absolute value sign surrounding $x_k$, so when you take the derivative what should pop out is $2|x_j|$ not just $2x_j$, correct me if I am wrong – Carlos - the Mongoose - Danger Oct 23 '15 at 17:46
• @Lookbehindyou well |t|^2 = t^2 for any $t\in\Bbb R$. Moreover, note that $\frac{d}{t}|t|^2 = 2|t|\operatorname{sign}(t)=2t$, where $\operatorname{sign}$ is the sign function – Surb Feb 29 '16 at 14:49
• Could you please explain what happened to the sigma in the last step? – Gigili Aug 16 '16 at 12:38
• @Gigili: If $k\neq j$, then $\frac{\partial }{\partial x_j}x_k^2=0$. Then, $$\sum_{k=1}^n\frac{\partial }{\partial x_j}x_k^2=0+...+0+2x_j+0+...+0=2x_j.$$ – Surb Aug 16 '16 at 12:44

Another approach that extends to more general settings is to use the connection between the norm and the inner product, $$\|x\|^2 = (x,x).$$

We have the finite difference, \begin{align} \|x+sh\|^2 - \|x\|^2 &= (x+sh,x+sh) - (x,x) \\ &= (x,x) + 2s(x,h) + s^2(h,h) - (x,x) \\ &= 2s(x,h) + s^2(h,h). \end{align}

The gradient acting in the direction $$h$$ is the limit of this finite difference as the stepsize goes to zero, \begin{align} (\nabla\|x\|^2, h) &:= \lim_{s \rightarrow 0} \frac{1}{s}\left[\|x+sh\|^2 - \|x\|^2\right] \\ &= \lim_{s \rightarrow 0} \frac{1}{s}\left[2s(x,h) + s^2(h,h)\right] \\ &= (2x,h). \end{align} Since this holds for any direction $$h$$, the gradient must be $$\nabla \|x\|^2 = 2x$$.