Subspace of Vectors orthogonal to every vectors in a given Vector space

Question

In a vector space V(dim-n), prove that the set of all vectors orthogonal to any vector( not equal to 0) form a subspace V[dim: (n-1)].

I am wondering how the n-1 is coming in the in the picture?

Answer 1

It is possible to prove this without the rank-nullity theorem. Let $V$ be our $n$-dimensional space, and suppose $w \in V$ is non-zero. Let $$W = \{w\}^\perp = \{v \in V : w \cdot v = 0\}.$$ Since you asked specifically where the $n - 1$ comes from, I'm not going to show that $W$ is a subspace.

The way we shall prove it is to show that, given any basis $B$ for $W$, that $B \cup \{w\}$ is a basis for $V$. $V$ is $n$-dimensional, meaning that $B \cup \{w\}$ must therefore contain $n$ vectors. Note that $w \cdot w = \|w\|^2 > 0$ as $w \neq 0$, hence $w \notin W$. We therefore have $w \notin B$, and so $B$ must only contain $n - 1$ vectors.

Suppose $B = \{u_1, \ldots, u_k\}$ is a basis for $W$. Since $w \notin W = \operatorname{span}\{u_1, \ldots, u_k\}$, we therefore have $B \cup \{w\}$ is linearly independent. So, we simply need to show that $B \cup \{w\}$ is spanning.

Suppose that $v \in V$. We need to write $V$ as a linear combination of vectors from $B \cup \{w\}$. Let $$u = v - \frac{v \cdot w}{w \cdot w}w.$$ Then $u \in W$, as $$u \cdot w = \left(v - \frac{v \cdot w}{w \cdot w}w\right) \cdot w = v\cdot w - \frac{v \cdot w}{w \cdot w} w \cdot w = 0.$$ Since $B$ spans $W$, there exist $a_1, \ldots, a_k$ such that $$a_1 u_1 + \ldots + a_k u_k = u = v - \frac{v \cdot w}{w \cdot w}w.$$ But then, rearranging, $$a_1 u_1 + \ldots + a_k u_k + \frac{v \cdot w}{w \cdot w}w = v.$$ The left hand side is a linear combination of $B \cup \{w\}$, proving $B \cup \{w\}$ is indeed a basis for $V$ and contains $n$ vectors. Thus, $B$ contains $n - 1$ vectors.