Dimension of vector of subspace inner product

Question

Suppose $v \in \mathbb R^2, v \ne 0.$ Then the subspace $\{x \in \mathbb R^2, x \cdot v = 0\}$ is of dimension
a) -1
b) 1
c) 2
d) 0.

Note that $\cdot$ denotes the inner product of two vectors in $\mathbb R^2$.

Answer 1

Let $v=(v_1 , v_2$), $v_i \neq 0$

Let $x=(y,z$)

$x.v = y v_1 + z v_2 = 0$

This is one equation in two unknowns. Hope you can take from here

Answer 2

Hints:

• Can a subspace have dimension $-1$?
• If the dimension is $2$, then $v$ is orthogonal to itself.
• If the dimension is $0$, then $v$ has no nonzero orthogonal vector.

Answer 3

Let $U=\{tv: t \in \mathbb R\}$, then $\{x \in \mathbb R^2: x \cdot v=0\}=U^{\perp}$ and $\mathbb R^2= U \oplus U^{\perp}$.

Can you proceed ?