Determining if a linear transformation across abstract vector spaces is surjective given a non-square coefficient matrix I have been having a bit of trouble with this problem. It can be viewed here, at the very bottom, question C23, the answer is provided but obviously not in enough detail. Here is the definition of the linear transformation. I understand that I must get $\vec{v}$ in terms of $\vec{u}$ but I am unsure of how to do that.
$$
T:\mathbb{C}^5 \longrightarrow P_3 \qquad
T\left(
\begin{bmatrix}
a \\ b \\ c \\ d \\ e
\end{bmatrix}
\right) =
a + (b+c)x + (c+d)x^2 + (d+e)x^3
$$
All of the other examples, namely "Example SAV" (see link), have just involved taking the inverse of a coefficient matrix when the equation is of the form $A\vec{u}=\vec{v}$. However, the coefficient matrix in this example is not square, it is rectangular provided below. How do I prove that this transformation is surjective?
$$
\begin{bmatrix}
1&0&0&0&0\\
0&1&1&0&0\\
0&0&1&1&0\\
0&0&0&1&1
\end{bmatrix}
$$
 A: Surjective means that the image of the transformation $T$ (also called the range) is all of the target vector space. In coordinates (meaning that you have explicitly chosen a basis for the domain $\mathbb{C}^5$ and the codomain $P_3$), you want the column space (span of columns of the matrix) to be all of $\mathbb{C}^4$.
In order to verify the latter, we apply the Gaussian elimination algorithm, which converts
$$
A = 
\begin{bmatrix}
1&0&0&0&0\\
0&1&1&0&0\\
0&0&1&1&0\\
0&0&0&1&1
\end{bmatrix}
$$
to the reduced row-echelon form
$$
B = 
\begin{bmatrix}
1&0&0&0&0\\
0&1&0&0&1\\
0&0&1&0&-1\\
0&0&0&1&1
\end{bmatrix}.
$$
Since $B$ has a pivot in each row, the transformation is onto. In fact, we can say more: the columns with pivots, namely $\{1, 2, 3, 4\}$, are the same columns in the original matrix who span the image of the transformation. In other words, with the notation $\mathbf{e}_j \in \mathbb{C}^5$ for the standard basis vector with $1$ in coordinate $j$ and $0$ elsewhere,
$$
\bigl\{ T(\mathbf{e}_1), T(\mathbf{e}_2), T(\mathbf{e}_3), T(\mathbf{e}_4) \bigr\} 
= \bigl\{ 1, x, x + x^2, x^2 + x^3 \bigr\}
$$
spans $P_3$.

Edited to respond to comments.
Working directly from the definition of surjective, given the transformation $T: \mathbb{C}^5 \to P_3$ and arbitrary $p \in P_3$, we must find $\mathbf{v} \in \mathbb{C}^5$ such that $T(\mathbf{v}) = p$. Start by naming coefficients for $p$:
$$
p = q + rx + sx^2 + tx^3.
$$
Since
$$
T \left( \begin{bmatrix} a\\b\\c\\d\\e \end{bmatrix} \right) 
= a + (b+c)\, x + (c+d)\, x^2 + (d+e)\, x^3,
$$
by equating coefficients, we have the system of linear equations
$$
\left\{
\begin{array}{rcrcrcrcrcr}
a &  &  &  &  &  &  &  &  &= &q \\
  &  &b &+ &c &  &  &  &  &= &r \\
  &  &  &  &c &+ &d &  &  &= &s \\
  &  &  &  &  &  &d &+ &e &= &t
\end{array}
\right.
$$
which can be represented as the augmented linear system
$$
\left[
\begin{matrix}
1&0&0&0&0\\
0&1&1&0&0\\
0&0&1&1&0\\
0&0&0&1&1
\end{matrix}
\;\middle|\;
\begin{matrix}
q\\r\\s\\t
\end{matrix}
\right].
$$
Row reduction yields
$$
\left[
\begin{matrix}
1&0&0&0&0\\
0&1&0&0&1\\
0&0&1&0&-1\\
0&0&0&1&1
\end{matrix}
\;\middle|\;
\begin{matrix}
q\\r-s+t\\s-t\\t
\end{matrix}
\right]. 
$$
Since the $5$th column has no pivot, the variable $e$ is free. Thus, we have
$$
\begin{bmatrix} a\\b\\c\\d\\e \end{bmatrix}
= \begin{bmatrix} q\\r-s+t-e\\s-t+e\\t-e\\e \end{bmatrix}
= \begin{bmatrix} q\\r-s+t\\s-t\\t\\0 \end{bmatrix} 
+ e \begin{bmatrix} 0\\-1\\1\\-1\\1 \end{bmatrix},
$$
for any $e \in \mathbb{C}$. All we need to establish surjectivity is at least one solution. Here we’ve parametrized an (infinite) $1$-dimensional family of solutions to the equation $T(\mathbf{v}) = p$.
You should plug this $\mathbf{v}$ into $T$ and check that it always produces $p$.
