I have the following claim: $\DeclareMathOperator{\Span}{Span}$
Let $V$ a vector space over the field $\mathbb{F}$.
Need to prove that for each $\vec{u},\vec{v},\vec{w}\in V$ such that:$$\Span(\vec{u},\vec{v},\vec{w})=\Span(\vec{u},\vec{u}+\vec{v},\vec{u}+\vec{v}+\vec{w})$$
My attempt:
$Proof.$ Let $V$ a vector space over the field $\mathbb{F}$.
By definition we have:
$$\Span(\vec{u},\vec{v},\vec{w})=\{\beta_1\vec{u}+\beta_2\vec{v}+\beta_3\vec{w}:\beta_1,\beta_2,\beta_3\in \mathbb{F} \}$$
In addition, by $\Span(\vec{u},\vec{u}+\vec{v},\vec{u}+\vec{v}+\vec{w})$ definition we get:
\begin{equation}
\begin{split}
\Span(\vec{u},\vec{u}+\vec{v},\vec{u}+\vec{v}+\vec{w}) &=\{\alpha_1\vec{u}+(\vec{u}+\vec{v})\alpha_2+(\vec{u}+\vec{v}+\vec{w})\alpha_3 :\alpha_1,\alpha_2,\alpha_3 \in \mathbb{F} \} \\&= \{ \alpha_1\vec{u}+\alpha_2\vec{u}+\alpha_2\vec{v}+\alpha_3\vec{u}+\alpha_3\vec{u}+\alpha_3\vec{v}\}
\\&=\{(\alpha_1+\alpha_2+\alpha_3)\vec{u}+(\alpha_2+\alpha_3)\vec{v}+\alpha_3\vec{w}:\alpha_1,\alpha_2,\alpha_3 \in \mathbb{F} \}
\\&=\{\beta_1\vec{u}+\beta_2\vec{v}+\beta_3\vec{w}:\beta_1,\beta_2,\beta_3\in \mathbb{F}\}
\\&=\Span{(\vec{u},\vec{v},\vec{w})} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \blacksquare
\end{split}
\end{equation}
I know that in order to show equality between two sets, I have to apply double inclusion.
However, any step in my proof has equality. Thus, I will be glad to see what you think about the proof, whether she is correct, or not, and if not to elaborate. Thanks!
Best Answer
I suggest another proof, using matrices:$\DeclareMathOperator{\Span}{Span}$
It is obvious that $\vec u,\: \vec u+\vec v, \:\vec u+\vec v+\vec w$ belong to $\Span(\vec u, \vec v,\vec w)$ so that $\;\Span(\vec u,\vec u+\vec v, \vec u+\vec v+\vec w)\subseteq\Span(\vec u, \vec v,\vec w)$.
Therefore all that has to be proved is the reverse inclusion. For this, observe we can write $$\begin{bmatrix}\vec u\\\vec u+\vec v\\ \vec u+\vec v+\vec w\end{bmatrix}= \underbrace{\begin{bmatrix}1&0&0 \\1&1&0\\1&1&1\end{bmatrix}}_{A}\begin{bmatrix}\vec u\\\vec v \\ \vec w\end{bmatrix}$$ The matrix $A$ is invertible, so we deduce $$\begin{bmatrix}\vec u\\\vec v \\ \vec w\end{bmatrix}=A^{-1}\begin{bmatrix}\vec u\\\vec u+\vec v\\ \vec u+\vec v+\vec w\end{bmatrix}=\begin{bmatrix}1&0&0\\-1&1&0\\ 0&-1&1\end{bmatrix}\begin{bmatrix}\vec u\\\vec u+\vec v\\ \vec u+\vec v+\vec w \end{bmatrix},$$ which shows the reverse inclusion.