What does $\left\Vert \mathbf{u}\right\Vert$ mean in this equation? How would this equation be performed? I'm extremely terrible in discrete mathematics and a simplistic answer would be ideal. (Don't answer it directly, I want to practice)

enter image description here

  • 3
    $\begingroup$ It looks like you have to calculate the norm of the given vector. $\endgroup$ – Hirshy Aug 26 '15 at 8:05

The norm of a vector $(1,3,4,11,13)$ is $\sqrt{1^2+3^2+4^2+11^2+13^2}$. It is an extension of Pythagoras' Theorem.

  • 8
    $\begingroup$ For this you're assuming that we have the euclidean norm. It is at least worth mentioning that there are other norms which one has to deal with in a different way. $\endgroup$ – Hirshy Aug 26 '15 at 8:08
  • $\begingroup$ Surprisingly this was the solution. Thank you Michael. $\endgroup$ – Oliver K Aug 26 '15 at 8:34
  • $\begingroup$ Is this what people mean when they call it the "Euclidian (L2) norm"? $\endgroup$ – Mark White Nov 30 '18 at 18:37
  • $\begingroup$ Yes, that's right. $\endgroup$ – Empy2 Nov 30 '18 at 18:58
  • $\begingroup$ @Hirshy what are the other kind of norms different from this one? $\endgroup$ – bikalpa Feb 26 '19 at 13:44

To elaborate on my comment on Michael's answer:

The symbol $\left\Vert\mathbf{u}\right\Vert$ for a vetor $\mathbf{u}$ usually stands for the norm of that vector. A norm is "a function that assigns a strictly positive length or size to each vector in a vector space" (quoted from wikipedia).

Having a normed vector space enables you to talk about e.g. the length of a vector. A common example would be the vector space $\mathbb R^n$ with the euclidean norm which is the norm induced by the dot product $\langle v,w\rangle = \sum\limits_{i=1}^n v_iw_i$ where $v=\begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix},w=\begin{pmatrix} w_1 \\ \dots \\ w_n\end{pmatrix}$, but there are other norms and inner products one can use.

Let $\mathcal V$ be a vector space with $\dim(\mathcal V)=n$ over $\mathbb R$ and $\langle\cdot,\cdot\rangle$ an inner product on $\mathcal V$. Then $(\mathcal V,\left\Vert\cdot\right\Vert)$ is a normed vector space with $\left\Vert v\right\Vert:= \sqrt{\langle v,v\rangle}$. The norm of a vector can be interpreted as the length of the vector but it is dependent on which inner product one uses.

  • 1
    $\begingroup$ Thanks Hirshy. A great informative post. +1 $\endgroup$ – Oliver K Aug 26 '15 at 9:59

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.