How to show 2 spanning sets are equal? Suppose that $\{u, v, w\}$ is a linearly independent set of vectors in $\mathbb{R}^{50}$.
Show that Span$\{u, v, w\}$ = Span$\{u − v, u − 2v + w, v + w\}$.

I know that in order to do this, I need to prove that each span is a subset of the other. So far, I have: $A :=$ Span$\{u, v, w\}$ and $B := $Span$\{u − v, u − 2v + w, v + w\}$.
Then, I showed that:


$u-v = -1(u)-1(v)+0(w)$ so $b1 ∈ A$.
$u-2v+w = 1(u)-2(v)+1(w)$ so $b2 ∈ A$.
$v+w = 0(u)+1(v)+1(w)$ so $b3 ∈ A$. 


Since any element $b$ in $B$ can be written as a linear combination of $b1, b2$, and $b3, B$ is a subset of $A$. I struggle with how to prove that $A$ is a subset of B, though. Guess and check is becoming very tedious.
Hint: solve some systems of linear equations to express each of $u, v, w$ as linear combinations of the second set of vectors.
It's easy to observe that $\mathrm{span}B \subseteq \mathrm{span}A$.

Let $\alpha=\{u,v,w\}$, $\beta=\{u-v,u-2v,v+w\}$.

The matrix for the change of coordinate from $\beta$ to $\alpha$ (we don't write from $\alpha$ to $\beta$ since it's more difficult) is

$$
[{\sf Id}]_\beta^\alpha =
\begin{bmatrix}
1 &amp; 1 &amp; 0 \\ -1 &amp; -2 &amp; 1 \\ 0 &amp; 0 &amp; 1
\end{bmatrix}.
$$

Observe that this matrix is invertible since it has linearly independent columns, so ${\sf Id}: \mathrm{span}B \to \mathrm{span}A$ is bijective.