How to fully describe a set consting of orthogonal to v

Question

Let $\hat v = \begin{bmatrix}a\\b\end{bmatrix}$. Describe the set $H$ consisting of vectors orthogonal to $\hat v$. (Consider $\hat v=\hat 0$ and $\hat v \neq \hat 0$)

My answer which I only received 2/10 points for was:

If H is orthogonal to $\hat v = \begin{bmatrix}a\\b\end{bmatrix}$ then this implies $\hat v H = \hat 0$ when $\hat v=\hat 0$ and $\hat v \neq \hat 0$. If $\hat v \neq \hat 0$ then set H would have to be perpendicular to in fact be orthogonal to $\hat v$ because for a pair of vectors to be orthogonal their dot product must be zero. For example, $\begin{bmatrix}2\\3\end{bmatrix} *\begin{bmatrix}-6\\4\end{bmatrix}=0$

Answer 1

Essentially you are asked to solve the linear equation:

$$ax+by=0\tag{1}$$

given $a$ and $b$. The set $H$ is the set of solutions to (1).

• If $a=b=0$, can you show that $H=\mathbb{R}^2$?
• If $(a,b)\neq (0,0)$, then without loss of generality suppose $a\neq 0$. What can you tell about (1)?

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To analyze your solution:

If H is orthogonal to $\hat v = \begin{bmatrix}a\\b\end{bmatrix}$ then this implies $\hat v H = \hat 0$ when $\hat v=\hat 0$ and $\hat v \neq \hat 0$.

This says nothing about $H$ but repeating the definition of orthogonality. Also $v=0$ and $v\neq 0$ cannot be true at the same time.

If $\hat v \neq \hat 0$ then set H would have to be perpendicular to in fact be orthogonal to $\hat v$ because for a pair of vectors to be orthogonal their dot product must be zero. For example, $\begin{bmatrix}2\\3\end{bmatrix} *\begin{bmatrix}-6\\4\end{bmatrix}=0$

Again, you says nothing about $H$ but giving an element of $H$ for a particular $v$.

Answer 2

Generally, we talk about vectors orthogonal to a subspace. So, let $A= \text{span}(\hat v)$. Then, we can talk about all vectors orthogonal to $H$ (and thus $\hat v$) as: $$A^\perp = \lbrace \hat x\in V\mid \hat x\cdot \hat v = 0\rbrace$$ Here, I'm denoting our vector space as $V$.

The set (and in fact subspace) $A^\perp$ should be the set $H$ you're looking for.

There are a few issues with your answer:

In general, $\hat v H$ isn't the dot product. Moreover, $H$ is a set, not a specific vector, so we don't have a well defined matrix multiplication here.

If we have two vectors $\hat a = \begin{pmatrix} a_1 \\ a_2\end{pmatrix}$ and $\hat b = \begin{pmatrix} b_1 \\ b_2\end{pmatrix}$, we can denote their dot product not as $\hat a\hat b$, but as $\hat a^t\hat b$, where $\hat a^t$ is the transpose. Writing this out we see this is the correct definition: $$\hat a^t\hat b = \begin{pmatrix} a_1 & a_2\end{pmatrix}\begin{pmatrix} b_1 \\ b_2\end{pmatrix} = a_1b_1+a_2b_2 = \hat a\cdot\hat b$$