How to compute the orthogonal complement of a matrix given an inner product.

Question

Suppose $v_1 = \begin{bmatrix} -4 \\ 1 \\ 2 \end{bmatrix}$, $v_2 = \begin{bmatrix} 0 \\ -1 \\ 2 \end{bmatrix}, S = \{v_1, v_2\}$, and \begin{align*}\langle x,y \rangle := a_1b_1 + 4a_2b_2 + a_3b_3\end{align*} is an inner product on $\mathbb{R}^3$, compute for a basis of $S^{\bot}$.

According to my understanding, $S^{\bot}$, also known as the orthogonal complement, is basically the null space (Am I right on that?), so I want to know if my solution and conclusion is correct:

First, I performed elementary row operations to get the RREF: \begin{align*} \text{RREF: } \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} \end{align*} Solving the matrix equation: \begin{align*} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \end{align*} Since the system has a unique solution, the null space is $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$. Hence, $S^{\bot}$ has no basis.

Attempt to Fred's suggestion: \begin{align*} \text{Let }v &= \begin{bmatrix} x \\ y \\ z \end{bmatrix}\\\\ \langle v, v_1 \rangle &= x(-4) + 4y(1) + z(2)\\ &= -4x + 4y + 2z\\\\ \langle v, v_2 \rangle &= x(0) + 4y(-1) + z(2)\\ &= -4y + 2z\\\\ \langle v, v_1 \rangle &= \langle v, v_2\rangle\\ -4x+4y+2z &= -4y+2z\\ -4x+8y &= 0\\\\ \text{Let }x = 1&:\\ -4(1) + 8y &= 0\\ -4 + 8y &= 0\\ y &= \dfrac{1}{2}\\\\ -4y + 2z &= 0\\ -2 + 2z &= 0\\ z &= 1 \end{align*} Hence, $v = \begin{bmatrix} 1 \\ 1/2 \\ 1 \end{bmatrix}$ is a basis of $S^{\bot}$

Answer 1

We have that $ \dim S=2$ and

$$ \mathbb R^3=S \oplus S^{\perp}.$$

Hence $ \dim S^{\perp}=1.$ This gives that $S^{\perp}= \{\alpha v: \alpha \in \mathbb R^3\}$ with $v \ne 0.$

Hence you have to determine a vector $v \in \mathbb R^3$ such that $v \ne 0$ and $\langle v,v_1 \rangle = \langle v,v_2 \rangle=0.$