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

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 $$S^{\bot}$$

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 $$\langle v,v_1 \rangle = \langle v,v_2 \rangle=0.$$

• What do you mean by $v \neq 0$?
– muw
May 19, 2021 at 13:39
• I mean $v \ne (0,0,0)^T.$
– Fred
May 19, 2021 at 13:41
• I understand what you are saying. I have to find $v$ such that $\langle v,v_1\rangle$ and $\langle v,v_2\rangle$ are both equal to 0, without $v$ being equal to $\begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$. Can you guide me as to how to do that? I understand the process but I have no clue how to get $v$.
– muw
May 19, 2021 at 14:39
• Hi, please check my post, I updated it with my attempt. Is my understanding correct?
– muw
May 19, 2021 at 15:08
• That is correct. Jun 19, 2021 at 16:25