Question 1160969: Let u=[1, 0, -1, 3], v=[0, 1, 4, 4], and let W the subspace of R^4 spanned by {u,v}. Find a basis for W⊥.
Answer by CPhill(2285)   (Show Source):
You can put this solution on YOUR website! To find a basis for $W^\perp$ (the orthogonal complement of $W$), we need to find all vectors $x = [x_1, x_2, x_3, x_4]^T \in \mathbb{R}^4$ that are orthogonal to both basis vectors $u$ and $v$.
This requires solving the system:
$$u \cdot x = 0 \implies x_1 - x_3 + 3x_4 = 0$$
$$v \cdot x = 0 \implies x_2 + 4x_3 + 4x_4 = 0$$
Writing this system in augmented matrix form:
$$\begin{bmatrix} 1 & 0 & -1 & 3 & \vert{} & 0 \\ 0 & 1 & 4 & 4 & \vert{} & 0 \end{bmatrix}$$
This matrix is already in **reduced row echelon form (RREF)**.
### Step 1: Identify Free and Pivot Variables
* **Pivot variables:** $x_1$ and $x_2$
* **Free variables:** $x_3$ and $x_4$
Expressing the pivot variables in terms of the free variables $x_3$ and $x_4$:
$$x_1 = x_3 - 3x_4$$
$$x_2 = -4x_3 - 4x_4$$
### Step 2: Write the General Solution Vector
$$x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} x_3 - 3x_4 \\ -4x_3 - 4x_4 \\ x_3 \\ x_4 \end{bmatrix} = x_3 \begin{bmatrix} 1 \\ -4 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -3 \\ -4 \\ 0 \\ 1 \end{bmatrix}$$
### Step 3: Extract the Basis
Setting $x_3 = 1, x_4 = 0$ gives the first basis vector, and setting $x_3 = 0, x_4 = 1$ gives the second:
$$\text{Basis for } W^\perp = \left\{ \begin{bmatrix} 1 \\ -4 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ -4 \\ 0 \\ 1 \end{bmatrix} \right\}$$
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