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?

Answer 1

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

$$v_2=v-v_1$$

Answer 2

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 $w_2$.