Linear algebra - proving two spans are equal There's a problem in a homework assignment I believe I've solved and proved but I want to be sure:
Let $V$ be a linear vector space. If $u,v,w,t$ are vectors in $V$, $Sp\{ u+v-3w\ ,\ 2v-w\ ,\ w+t\ ,\ v+w \}=Sp\{u,v,w,t\}$ .
I believe this statement is correct because if we were to approach the members in each group as matrices, we'll be able to use row operations to take the group on the left and transform it into the group on the right like this:
$$
\left\{ u+v-3w\ ,\ 2v-w\ ,\ w+t\ ,\ v+w \right\} =
\left\{
\begin{matrix}
u&+v&-3w&+0t\\
0u&+2v&-w&+0t\\
0u&+0v&+w&+t\\
0u&+v&+w&+0t\\
\end{matrix}
\right.
\ 
\Rightarrow
\ 
\begin{pmatrix}
1&1&-3&0\\0&2&-1&0\\0&0&1&1\\0&1&1&0
\end{pmatrix}
\\\ \\
\rightarrow
\begin{pmatrix}
1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1
\end{pmatrix}
=\{u,v,w,t\}
$$
Since one group can be transformed into another using linear combinations of its vectors, their spans must be the same.
Am I correct in this assertion?
 A: Your solution is correct, but it is worth seeing why it works in a more rigorous manner. You want to show that
$$ \mathrm{span} \{ u, v, w, t \} = \mathrm{span} \{ u + v - 3w, 2v - w, w + t, v + w \}. $$
Let's start with the left hand side and show that it belongs to the right hand side. A general element $x$ in $\mathrm{span} \{ u, v, w, t \}$ is an element of the form $x = au + bv + cw + dt$ where $a,b,c,d \in \mathbb{R}$. A general element in $\mathrm{span} \{ u + v - 3w, 2v - w, w + t, v + w \}$ is of the form 
$$a'(u + v - 3w) + b'(2v - w) + c'(w + t) + d'(v + w). $$
To show that $x$ belongs to the right hand side, we need to find $a',b',c',d'$ such that
$$ x = au + bv + cw + dt = a'(u + v - 3w) + b'(2v - w) + c'(w + t) + d'(v + w) = a'u + (a' + 2b' + d')v + (c' + d' - b' - 3a')w + c' t. $$
By comparing coefficient, we see that is is enough to find $a',b',c',d'$ such that
$$ a' = a, \\
  a' + 2b' + d' = b, \\
  c' + d' - b' - 3a' = c, \\
  c' = d.
$$
Written in matrix form, we have the system of equations
$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 
  1 & 2 & 0 & 1 \\
  -3 & -1 & 1 & 1 \\
  0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} a' \\ b' \\ c' \\ d' \end{pmatrix} = A \begin{pmatrix} a' \\ b' \\ c' \\ d' \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}. $$
This system of equations will have a solution for all $a,b,c,d$ if and only if the matrix $A$ will be of full rank which can be checked by performing elementary operations on $A$ like you did.
