If $\vec{w}$ can be expressed as linearcombination of $\vec{u}$ and $\vec{v}$, then does it imply that $\vec{u}$ can be expressed as linearcombination of $\vec{v}$ and $\vec{w}$?
Attempt to solve
Statement says:
$$ \vec{w} = a \vec{u} + b \vec{v} \implies \vec{u}=c\vec{v}+d\vec{w} $$ when $\{a,b,c,d\} \in \mathbb{R}$ and $\{\vec{w},\vec{v},\vec{u}\}\in \mathbb{R}^n$
I could rewrite $\vec{u}$ as:
$$ u = \frac{1}{a}\vec{w} + \frac{b}{a}\vec{v} \implies d = \frac{b}{a}, c = \frac{1}{a} $$
$$ \implies \vec{u} = \frac{1}{a}(\vec{w} + b\vec{v}) $$
Now is span of $\vec{u}$ all $\mathbb{R}^n$ ? If span of $\vec{u}$ is all $\mathbb{R}^n$ it means statement is true otherwise is not.
it would seem we are limited to a line since we are scaling both vectors with $\frac{1}{a}$ but not quite sure about this.
Consider $w=v=(1,0)$ and $u=(0,1)$. Clearly $w$ is a linear combination of $u,v$, being $0u+1v$. However, the span of $v,w$ is still only $k(1,0)$ for $k\in\mathbb R$, and this span does not include $u$. Thus the statement is false.