I'm stuck on the following problem, any help is appreciated
Here it is:
Let S be the subspace of $ℝ^4$ spanned by $x_1=(1,0,-2,1)^T$ and $x_2=0,1,3,-2)^T$ Find a basis for $S^⊥$
I know that is is the set of all vectors perpendicular to a vector, at least that's the case when S is spanned by 1 vector. Now that it's spanned by 2 I don'tknow how to solve this question.
Thanks in advance :)
You are looking for a basis of $S^\perp$, which is defined as
$$S^\perp := \{y\in \mathbb{R}^4: x_1\cdot y = x_2 \cdot y = 0\}.$$
Therefore, some vector $y\in \mathbb{R}^4$ is contained in $S^\perp$ if and only if
$$\begin{bmatrix} x_1^T\\x_2^T\end{bmatrix}y = \begin{bmatrix} 1&0&−2&1\\0&1&3&-2\end{bmatrix}y = \begin{bmatrix}0\\0\end{bmatrix}.$$
We can use Gaussian elimination (see https://en.wikipedia.org/wiki/Gaussian_elimination or Gaussian Elimination General Solution) to solve this system of equations. The answer we get from Gaussian elimination will have two variables `free variables' (see the second link provided above) that will give us a basis.
The way of thinking is almost the same. Any vector of $S$ has the form $y=ax_1+bx_2$. what we want is ^the set of vectors s.t. are perpendicular to any y, or in math words $$\lbrace z\in \mathbb{R}^4:z\dot{}(ax_1+bx_2)=0,\forall a,b\in\mathbb{R}\rbrace$$ Then just characterize the set.
Search for a vector $x_3 = (x, y, z, t)$ such that <$x_1,x_3$>$=0$ and <$x_2,x_3$>$=0$.
$$\begin{cases} x-2z+t=0 \\ y+3z-2t=0 \end{cases} $$
A vector $x_3$ satisfying these equations could be $(1,-1,1,1)$.
Now find $x_4=(a, b, c, d)$ in the same way, $x_4$ must be such that <$x_4,x_1$>$=0$,<$x_4,x_2$>$=0$,<$x_4,x_3$>$=0$.
$$\begin{cases} a-2c+d=0 \\ b+3c-2d=0 \\ a-b+c+d=0 \end{cases} $$