How to show 2 spanning sets are equal?

Question

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.

Answer 1

Hint: solve some systems of linear equations to express each of $u, v, w$ as linear combinations of the second set of vectors.

Answer 2

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

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

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