Find the coordinates for $v$ in the subspace $W$ spanned by the following vectors: 
I'm confused as to how I solve this. I was told to project each $u$ onto $v$ but how would that get me to the answer that's required?
 A: Basically what @PrahLad said, $v = a_1u_1 + a_2u_2 + a_3u_3$ So since $u_1,u_2,u_3$ are orthogonal you can use the formula $a = \frac{y \cdot u}{u \cdot u}$ or in your case: $\alpha = \frac{v \cdot u_i}{u_i\cdot u_i}$
So the first one for example will be 
$\frac{v \cdot u_1}{u_1 \cdot u_1} = $ (your calculation here)
A: Resolve the data vector into its components in the $u$ coordinate system. because the $u$ vectors are orthogonal, we can solve for the projection terms independently.

$$
\begin{array}{cccccccccc}
 v & = & 
\frac{\langle v,u_{1} \rangle}{\langle u_{1},u_{1} \rangle} & u_{1} & + & \frac{\langle v,u_{2} \rangle}{\langle u_{2},u_{2} \rangle} & u_{2} & + &
\frac{\langle v,u_{3} \rangle}{\langle u_{3},u_{3} \rangle} & u_{3} & \\
%
\left[
\begin{array}{r}
 5 \\ -2 \\ 9 \\ -5
\end{array}
\right]
& = &
 \frac{27}{22}  & \left[
\begin{array}{r}
 3 \\3 \\ 2 \\ 0
\end{array}
\right]
 & + & 
-\frac{11}{7}   & \left[
\begin{array}{r}
 2 \\ 0 \\ -3 \\ 1
\end{array}
\right]
 & + &
-\frac{65}{488} & \left[
\begin{array}{r}
 -6 \\ 2 \\ 6 \\ 30
\end{array}
\right] \\
%
&=&
\frac{1}{18788 }  & \left[
\begin{array}{r}
 -68799 \\ 101745 \\ -49419 \\ -10659
\end{array}
\right]
%
\end{array}
$$
A: Since $v \in W = \text{span}(u_1, u_2, u_3)$, there exist scalars $\alpha_1, \alpha_2, \alpha_3$ such that
$$v = \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3$$
Note that the scalars $\alpha_1, \alpha_2, \alpha_3$ are unique. (Why?) The scalars $\alpha_1, \alpha_2, \alpha_3$ are what we are asked to find.
For any $w \in W$, we have
$$(v,w) = \alpha_1 (u_1,w) + \alpha_2 (u_2,w) + \alpha_3 (u_3,w)$$
In particular, for $i = 1,2,3$, if we put $w = u_i$, we have
$$(v,u_i) = \alpha_1 (u_1,u_i) + \alpha_2 (u_2,u_i) + \alpha_3 (u_3,u_i) = \alpha_i (u_i,u_i)$$
since $u_1,u_2,u_3$ are orthogonal. Therefore $$\alpha_i = \frac{(v,u_i)}{(u_i,u_i)}$$
A: Try forming a matrix $A$ with the column vector $u_{1}$ as the first column, $u_{2}$ as the second, and $u_{3}$ as the third. Then, set up the equation $Ax=b$ where $b=v$ so that $Ax=v$ leaves you the task of solving for a $3\times 1$ column vector $x$. The components of this column vector $(x_{1}, x_{2}, x_{3})$ will be the respective coefficients in each of the boxes above. Does that make sense? I'm not sure that projection is required.
