Re: Orthogonal complement

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Originally Posted by

**Ulysses** Therefore $\displaystyle <x,y>=0 \rightarrow \xi_1 \eta_1+\xi_2 \eta_2=0 \rightarrow \frac{\xi_1}{\xi_2}=\frac{-\eta_2}{\eta_1}$ . And then $\displaystyle y=(\xi_2,-\xi_1)$.

We don't know if $\displaystyle \xi_2\neq 0$ . Better: $\displaystyle M^{\perp}=\{(\eta_1,\eta_2)\in\mathbb{R}^2:\eta_1 \xi_1+\eta_2\xi_2=0\}$ , then $\displaystyle \dim M^{\perp}=\dim \mathbb{R}^2-\textrm{rank}\;(\xi_1,\xi_2)=1$ and $\displaystyle (\xi_2,-\xi_1)$ is a non zero vector of $\displaystyle M^{\perp}$ . This means that $\displaystyle B=\{(\xi_2,-\xi_1)\}$ is a basis of $\displaystyle M^{\perp}$ and as a consequence $\displaystyle M^{\perp}=\textrm{Span}[(\xi_2,-\xi_1)]$

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b) A linearly independent set $\displaystyle \{x_1,x_2\}\subset{M}$

$\displaystyle B=\{x_1,x_2\}$ is a basis of $\displaystyle \mathbb{R}^2$ , so $\displaystyle M=\mathbb{R}^2$ and as a consequence $\displaystyle M^{\perp}=\{0\}$ .

Re: Orthogonal complement

Don't forget that an orthogonal complement is always a linear subspace. In (a), for example, $\displaystyle M^\perp$ cannot consist of a single vector. It will have to be the one-dimensional subspace consisting of all scalar multiples of that vector.

Your answer to (b) is essentially on the right lines, except that there is one vector that lies in any linear subspace. So the answer here will be the (zero-dimensional) subspace consisting of that one vector.