$$u1 = \begin{bmatrix} 1 \\ 2 \\ 1 \\ 1 \\ \end{bmatrix}\,, u2 = \begin{bmatrix} -2 \\ 1 \\ -1 \\ 1 \\ \end{bmatrix}\,, u3 = \begin{bmatrix} 1 \\ 1 \\ -2 \\ 1 \\ \end{bmatrix}\,, u4 = \begin{bmatrix} -1 \\ 1 \\ 1 \\ -2 \\ \end{bmatrix}\,, v = \begin{bmatrix} 4 \\ 5 \\ -3 \\ 3 \\ \end{bmatrix}$$

So if I'm interpretting this correctly I have to write v as v=c1p1 + c2p2 where p1 is span{u1} and p2 is span{u2, u3, u4}. How do I find a value for the span of just one vector if the span of it is that vector with any possible weight? What about for the span of those three vectors? And then I just add up those two spans?

  • $\begingroup$ No. Can you write down the definition of "span" here, please? $\endgroup$ – user296602 Apr 16 '18 at 21:05
  • $\begingroup$ The span $\langle \vec a\rangle$ for a vector $\vec a$ is the set of all vectors having the same direction as that of $\vec a$ but varying magnitude. Explicitly stated for $\Bbb R^3$ with the Euclidean metric, we have, $$\langle \vec a\rangle=\{\lambda\hat a\mid\lambda\in\Bbb R\}$$ $\endgroup$ – Prasun Biswas Apr 16 '18 at 21:22
  • $\begingroup$ @pythonhelpthrow Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Apr 19 '18 at 19:32

The span of a vector is the line (i.e. a subspace with dimension equal to $1$)

$$P(t)=t\begin{bmatrix} 1 \\ 2 \\ 1 \\ 1 \\ \end{bmatrix}$$

with $t\in \mathbb{R}$, that is more formally $$\operatorname{Span}(u) = \{t \cdot u : t \in \mathbb{R}\}$$

With reference with the new question after editing, yes we need to solve the linear system

$$x u_1+y u_2+x u_3+ wu_4=v$$

and then consider



  • $v_1=xu_1$
  • $v_2=y u_2+x u_3+ wu_4$

and since it is an orthogonal basis we can solve also by orthogonal projection, that is

$$x=\frac{u_1 \cdot v}{|u_1|}\implies v_1=x\frac{u_1}{|u_1|}=\frac{u_1 \cdot v}{|u_1|^2}u_1$$



  • $\begingroup$ So if it wants me to find the span of a vector I have to find a weight t? $\endgroup$ – pythonhelpthrow Apr 16 '18 at 21:08
  • $\begingroup$ The span of a set of vectors is, by definition, the set of all the (finite) linear combination of those vectors. In this case since we have one vector the span is the set of all the multiple of that vector. en.wikipedia.org/wiki/Linear_span $\endgroup$ – user Apr 16 '18 at 21:13
  • 1
    $\begingroup$ I think this answer is pretty misleading, especially for an asker who already thinks that the span is a single vector. It would be far better to write it as $\operatorname{Span}(u) = \{t \cdot u : t \in \mathbb{R}\}$. $\endgroup$ – user296602 Apr 16 '18 at 21:16
  • $\begingroup$ Thanks for the help I'm gonna edit the question to be more specific because I'm still confused as to how this applies to my problem $\endgroup$ – pythonhelpthrow Apr 16 '18 at 21:17
  • $\begingroup$ @user296602 Yes I think you are right, I add this notation to the answer. Thanks $\endgroup$ – user Apr 16 '18 at 21:18

The span of a set of vectors $\{v_1,\dots,v_m\}$ is the set of all linear combinations $a_1v_1+\cdots+a_mv_m$ of those vectors. If the set consists of only one vector, then its span is simply the set of all scalar multiples of that vector.

What you’re being asked to do in this problem, then, is to find two specific vectors $w_1=a_1u_1$ and $w_2=a_2u_2+a_3u_3+a_4u_4$ such that $v=w_1+w_2$. So, yes, once you’ve found the scalars that make $v=a_1u_1+a_2u_2+a_3u_3+a_4u_4$ you just add up the last three vectors.

Having gotten that out of the way, the condition that the $u$’s form an orthonormal basis is a big clue for how you might actually compute $w_1$ and $w_2$.

  • $\begingroup$ we can aslo use orthogonal projection, isn't it? $\endgroup$ – user Apr 16 '18 at 21:33
  • 1
    $\begingroup$ To actually compute $w_1$ and $w_2$, sure, but that wasn’t the question. Good point, though. $\endgroup$ – amd Apr 16 '18 at 21:38
  • $\begingroup$ Thanks! What about this OP math.stackexchange.com/q/2740287/505767? I've found a solution but maybe there are more strighforward ways? $\endgroup$ – user Apr 16 '18 at 21:40

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.