Finding the dimension and basis of an orthogonal space

Question

I am trying to find the basis and dimensions of the the space orthogonal $S$ which is in $\mathbb R^3$.

$$S = \begin{bmatrix}1\\2\\3\end{bmatrix}$$

So the dimension would be two because it is $3 - 1$.

The problem I am having is finding the basis. Would I do

$$ 1x+2y+3z = 0$$

Which would give me different basis. Am I on the correct track?

Answer 1

You just need find two independent vectors $[x~y~z]^T$ that satisfies $1x+2y+3z=0$.

For example, let $x_1=-3, y_1=0,z_1=1$, $\Rightarrow [x_1~y_1~z_1]^TS=-3\times 1+0\times 2+3\times 1=0$, also let $x_2=2, y_1=-1,z_1=0$, then $[x_2~y_2~z_2]^TS=2\times 1+(-1)\times 2+3\times 0=0$.

Therefore $\begin{bmatrix}x_1\\y_1\\z_1\end{bmatrix}=\begin{bmatrix}-3\\0\\1\end{bmatrix}$, and $\begin{bmatrix}x_2\\y_2\\z_2\end{bmatrix}=\begin{bmatrix}2\\-1\\0\end{bmatrix}$ are such two vectors, because they are also independent:

if $\alpha\begin{bmatrix}-3\\0\\1\end{bmatrix}+\beta\begin{bmatrix}2\\-1\\0\end{bmatrix}=0$, then $\begin{bmatrix}-3\alpha\\0\\\alpha\end{bmatrix}+\begin{bmatrix}2\beta\\-\beta\\0\end{bmatrix}=0$ , $$\Rightarrow \begin{bmatrix}-3\alpha+2\beta\\-\beta\\\alpha\end{bmatrix}=0$$, which implies $\alpha=\beta=0$

Answer 2

Well, the equation $1x+2y+3z=0$ indeed says that the inner product of vectors $S$ and $(x,y,z)^T$ is $0$, that is, they are orthogonal.

So we are looking for a basis among its solutions: e.g. you can take $y=1,\ z=0$ then $y=0,\ z=1$ and calculate the $x$'s for them.