<s>[INST] Linear transformations defined by $T(v) = Av$. Find all of possible $v$ I'm stuck on a problem. The problem is this:
The linear transformation $T : \Bbb{R}^4 \to \Bbb{R}^2$ is defined by $T(v) = Av$, where
$$A = \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix}$$
Find all vectors $v$ such that: $$T(v) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$
So I have forgotten how to do this. Do I:
- reduce $A$ to reduced row-echelon form (why do I do this? Is it because it's easy to solve once you have pivot columns and free variables)?
- rewrite the system of equations
Is this right:
\begin{align}
A &= \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & -5 & -2 & 7 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix} \\
&= \begin{bmatrix} 1 & 0 & \frac{1}{5} & \frac{-29}{5} \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix}
\end{align}
so: $v_4 = t, v_3 = s, v_2 = \frac{-2}{5}s - \frac{7}{5}t, v_1 = \frac{-1}{5}s + \frac{29}{5}t$
$$\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} \frac{-1}{5}s + \frac{29}{5}t \\ \frac{-2}{5}s - \frac{7}{5} t \\ s + 0t \\ 0 + t \end{bmatrix} = s \begin{bmatrix} \frac{-1}{5} \\ \frac{-2}{5} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} \frac{29}{5} \\ \frac{-7}{5} \\ 0 \\ 1 \end{bmatrix} $$
Is this the set of all $v$?
EDIT
I messed up, the first $\frac{7}{5}$ should be a $\frac{-7}{5}$
 [/INST] You found the solutions to the homogeneous system $Tv = 0$.
The solutions to the system $Tv = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ are of the form $$\left(\text{one particular solution to }Tv = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\right) \quad + \quad\left(\text{ any solution of the homogeneous system }\right)$$
We can guess one solution as $v = \begin{bmatrix} \frac45 \\ \frac35 \\ 0 \\ 0\end{bmatrix}$ so all solutions are of the form
$$v = \begin{bmatrix} \frac45 \\ \frac35 \\ 0 \\ 0\end{bmatrix} + s \begin{bmatrix} \frac{-1}{5} \\ \frac{-2}{5} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} \frac{29}{5} \\ \frac{-7}{5} \\ 0 \\ 1 \end{bmatrix}$$
for some $s,t \in \mathbb{R}$.
</s>