Orthogonal decomposition of vector v with respect to span W

Question

I need to find the orthogonal decomposition of vector V with respect to span W

$$ v= \begin{bmatrix} 2 \\ -1 \\ 5 \\ 6 \\ \end{bmatrix} $$ $$ w = span \begin{pmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -1 \\ 1 \end{bmatrix} \end{pmatrix} $$

I know that $ v = y + z $ where $ y \in W $ and $ z \in \text{orthogonal complement of W} $

Answer 1

Project $v$ onto $W$:

$$\begin{align} \text{proj}_W v &= \frac{1}{3} \cdot \left( \begin{bmatrix}2 & -1 & 5 & 6 \end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 1 \\ 0 \end{bmatrix} \right ) \begin{bmatrix}1 \\ 1 \\ 1 \\ 0 \end{bmatrix} + \frac{1}{3}\left( \begin{bmatrix}2 & -1 & 5 & 6 \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ -1 \\ 1 \end{bmatrix} \right ) \begin{bmatrix}1 \\ 0 \\ -1 \\ 1 \end{bmatrix} \\ &= \begin{bmatrix}2 \\ 2 \\ 2 \\ 0 \end{bmatrix} + \begin{bmatrix}1 \\ 0 \\ -1 \\ 1 \end{bmatrix} \\ &= \begin{bmatrix}3 \\ 2 \\ 1 \\ 1 \end{bmatrix} \end{align}$$

Then, $z$ is given by $z = v - \text{proj}_Wv$: $$\begin{align} z &= v - \text{proj}_Wv \\ &= \begin{bmatrix}2 \\ -1 \\ 5 \\ 6 \end{bmatrix} - \begin{bmatrix}3 \\ 2 \\ 1 \\ 1 \end{bmatrix} \\ &= \begin{bmatrix}-1 \\ -3 \\ 4 \\ 5 \end{bmatrix} \end{align}$$

$z$ is indeed orthogonal to $W$ since its dot product with both vectors in the spanning set for $W$ is 0.