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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.

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

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  • $\begingroup$ if I get u,v,w into matrix form, what do I augment it with? $\endgroup$ – AmaC Feb 28 '17 at 21:59
  • $\begingroup$ E.g. with $b_1 = u-v$, $b_2 = u-2v+w$, $b_3 = v+w$, $u = c_1 b_1 + c_2 b_2 + c_3 b_3$ would require $c_1 + c_2 = 1$, $-c_1 - 2 c_2 + c_3 = 0$, $c_2 + c_3 = 0$. $\endgroup$ – Robert Israel Feb 28 '17 at 22:43
  • $\begingroup$ @ Robert Isreal How did you get that c1+c2 = 1? And, why are the others equal to 0? Otherwise, I understand the rest $\endgroup$ – AmaC Feb 28 '17 at 22:59
  • $\begingroup$ $c_1 b_1 + c_2 b_2 + c_3 b_3 = (c_1 + c_2) u + (-c_1 - 2 c_1 + c_3) v + (c_2 + c_3) w$. You want this to be $u$. Since $u, v, w$ are linearly independent, the coefficient of $u$ must be $1$ and the coefficients of $v$ and $w$ must be $0$. $\endgroup$ – Robert Israel Mar 1 '17 at 0:23
  • $\begingroup$ Oh yes, this makes so much sense now, thank you very much for your knowledge and patience. $\endgroup$ – AmaC Mar 1 '17 at 2:11
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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.

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