Assume the set {u1...u4} is an orthogonal basis for R4. Write v as the sum of two vectors, one in span u1 and the other in span {u2, u3, u4}

Question

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

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?

No. Can you write down the definition of "span" here, please? — , Apr 16 '18 at 21:05
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, $$\langle \vec a\rangle={\lambda\hat a\mid\lambda\in\Bbb R}$$ — Prasun Biswas, Apr 16 '18 at 21:22
@pythonhelpthrow Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work — user, Apr 19 '18 at 19:32

user · Answer 1 · 2018-04-18T01:51:17.680

2

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

$$v=v_1+v_2$$

with

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

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

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

and

$$v_2=v-v_1$$

edited Apr 18 '18 at 01:51

answered Apr 16 '18 at 21:06

user

162,563

So if it wants me to find the span of a vector I have to find a weight t? – pythonhelpthrow Apr 16 '18 at 21:08
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. https://en.wikipedia.org/wiki/Linear_span – user Apr 16 '18 at 21:13
1

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 $\operatorname{Span}(u) = {t \cdot u : t \in \mathbb{R}}$. – Apr 16 '18 at 21:16
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 – pythonhelpthrow Apr 16 '18 at 21:17
@user296602 Yes I think you are right, I add this notation to the answer. Thanks – user Apr 16 '18 at 21:18
@user296602 My aim was to give a simple explanation of the concept by the idea of a "line", indicating that it is to be indented as a subspace of dimension $1$ in this case. – user Apr 16 '18 at 21:21
How can you solve for v1 without knowing the v that's in that equation? – pythonhelpthrow Apr 18 '18 at 00:55
v is the given vector then we find $x=\frac{14}{\sqrt 7}=\frac12 \sqrt 7$ thus $v_1=\frac12 \sqrt 7 u_1$ – user Apr 18 '18 at 01:02
Is that the absolute value of u1 or the norm of u1? – pythonhelpthrow Apr 18 '18 at 01:09
The norm of course, we are making a projection in order to find the component of $v$ on $u_1$. – user Apr 18 '18 at 01:12
Let us continue this discussion in chat. – pythonhelpthrow Apr 18 '18 at 01:13

amd · Answer 2 · 2018-04-16T21:40:09.173

1

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 $w_2$.

edited Apr 16 '18 at 21:40

answered Apr 16 '18 at 21:30

amd

55,082

we can aslo use orthogonal projection, isn't it? – user Apr 16 '18 at 21:33
1

To actually compute $w_1$ and $w_2$, sure, but that wasn’t the question. Good point, though. – amd Apr 16 '18 at 21:38
Thanks! What about this OP https://math.stackexchange.com/q/2740287/505767? I've found a solution but maybe there are more strighforward ways? – user Apr 16 '18 at 21:40

Assume the set {u1...u4} is an orthogonal basis for R4. Write v as the sum of two vectors, one in span u1 and the other in span {u2, u3, u4}

2 Answers2