Example 4.3.3.
Let \(v_1=(1,2,-1)\) and \(v_2=(3,1,2)\) and \(W = \{\alpha_1v_1+\alpha_2v_2:\alpha_1,\alpha_2\in \R\}\text{.}\) What is \(W\text{?}\) Can identify it geometrically? Yes, it is a plane passing through the origin. That the \(W\) can be written as \(W=\{(x,y,z):ax+by+cz=0\}\) for some \((a,b,c)\neq (0,0,0)\text{?}\) Can you find what are \(a,b,c\text{.}\)
From the concept of dot product, it easy to identify \((a,b,c)\) as a vector which is orthogonal/perpendicular to both \(v_1\) and \(v_2\text{.}\) In particular, we can find \((a,b,c)\) and \(v_1\times v_2\text{,}\) the cross product of \(v_1\) and \(v_2\text{.}\)
Suppose, we do not want to use the above concept to find \((a,b,c)\text{,}\) then what do we do?
Suppose \((x,y,z)\in W\text{,}\) Then there exists scalars \(\alpha\) and \(\beta\) such that
\begin{equation*}
\begin{pmatrix}x\\y\\z \end{pmatrix} =\alpha \begin{pmatrix}1\\2\\-1 \end{pmatrix} +\beta \begin{pmatrix}3\\1\\2 \end{pmatrix} = \begin{pmatrix}1 \amp 3\\2 \amp 1 \\-1 \amp 2 \end{pmatrix} \begin{pmatrix}\alpha\\\beta \end{pmatrix}\text{.}
\end{equation*}
In particular, \(W\) is the image space of \(\begin{pmatrix}1 \amp 3\\2 \amp 1 \\-1 \amp 2 \end{pmatrix}\text{.}\)
We need to find \((a,b,c)\) such that \(ax+by+cz=0\) for any \((x,y,z)\in W\text{.}\) In particular, we have
\begin{equation*}
a(\alpha+2\beta)+b(2\alpha+\beta)+c(-\alpha+2\beta)=0
\end{equation*}
for any \(\alpha,\beta\in \R\text{.}\) Note that \(\alpha,\beta\) is our choice and we can choose conveniently to find \(a,
b, c\text{.}\) It is easy to see that
\begin{align*}
a+2b-c \amp = \amp 0, \text{ for \(\alpha=1,\beta=0\) }\\
3a+b+2c \amp = \amp 0, \text{ for \(\alpha=0,\beta=1\) }
\end{align*}
This is same say substituting \(v_1,v_2\) in the equation \(ax=by=cz=0\text{.}\)
In particular, we have \(a,b,c\) such that
\begin{equation*}
\begin{pmatrix}1 \amp 2 \amp -1\\ 3 \amp 1 \amp 2 \end{pmatrix} \begin{pmatrix}a\\b\\c \end{pmatrix} =0\text{.}
\end{equation*}
Thus \((a,b,c)\) is the kernel of \(B=\begin{pmatrix}1 \amp 2 \amp -1\\ 3 \amp 1 \amp 2 \end{pmatrix}\) and \(W\) is the orthogonal complement of kernel of \(B\text{.}\)
Solving the above equations, we can find \(a=1,b=-1,c=-1\) as one of the choices. This implies \(W\) is the plane \(x-y+z=0\text{.}\)
