Let $X = \mathbb{R}^2$. Find the annihilator $M^{\perp}$ if $M$ is:
(a) $\{x\}$, where $x = (x_1,x_2) \not=0$.
(b) a linearly independent set $\{x_1,x_2\} \subseteq X$.
Here is what I tried so far. In both parts, I am stuck, not sure how to understand this fundamentally.
For part (a): \begin{align} M^{\perp}&=\{y \in X :y \perp M\} & \text{by definition of } M^\perp \\ &= \{y \in \mathbb{R}^2 : y \perp \{x\}\}\\ &= \{y \in \mathbb{R}^2 : \langle y,x \rangle=0\} \\ &= \{y \in \mathbb{R}^2 : y=c(-x_2,x_1)\} \end{align} and for part (b): \begin{align} M^{\perp}&=\{y \in X :y \perp M\} & \text{by definition of } M^\perp \\ &= \{y \in \mathbb{R}^2 : y \perp \mathbb{R}^2\}\\ &= \{y \in \mathbb{R}^2 : \langle y,\underbrace {\underset{\begin{subarray}{c} \text{linearly} \\ \text{independent} \end{subarray}}{\{x_1,x_2\}}}\rangle=0\} \\ &= \{y \in \mathbb{R}^2 : y=(0,0)\} \\ &= \{(0,0)\} \end{align}
Hint: Given a vector $v \in \mathbb{R}^2$ how can we find an orthogonal vector, i.e. if we draw this vector $v$ what must we do to find an orthogonal vector (i.e. what angle do we have to rotate through).
Hint:$^2$ Gram-Schmidt.
Hint:$^3$ If we have $n$ linearly independent vectors in $\mathbb{R}^n$ what can we say about the span of those vectors?