Find the general equation of the line of $span\left(\begin{bmatrix}2\\ -4\end{bmatrix}, \begin{bmatrix} 1\\
-2\end{bmatrix}\right)$
So I started with solving the matrix \begin{bmatrix}2 & -1 & 0\\ -4 & 2 & 0\end{bmatrix} which becomes \begin{bmatrix}1 & -\frac{1}{2} & 0\\ 0 & 0 & 0\end{bmatrix}
This shows that our vectors are linearly dependent since we have a parameter that can give us infinite values.
I took it apart into it's equation $x-\frac{1}{2}y=0$ and set y=t $\therefore$ $x-\frac{1}{2}t=0$.
We also have our vector equation $\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 2 \\ -4 \end{bmatrix}+ t\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ (This is probably where I messed up)
$x=2-t$, $y=-4+2t$
I'm pretty sure I can just use substitution from one of the equations, then solve for t, then just solve for x and y, but that doesn't seem right, and it makes solving the matrix useless. I can also try solving for x and y by substituting my parametric equations into $x-\frac{1}{2}y=0$ and solve for t, but then again, I'm not sure.
In the end I want to represent my equation as $Ax+By=0$
Best Answer
The vectors in your problem are in $\mathbb R^2.$ A subspace spanned by vectors in $\mathbb R^2$ could be the zero subspace, a line, or a plane. In this case, as you said, the two vectors are linearly dependent, and they aren't just the zero vector, so they span a line. You can easily see that the equation of that line is $2x-y=0$. There's no need for a parameter $t$ for a line in $2$ dimensions.