Which vectors in 3-dimensional space are linear combinations of $u$ and $v$ and also are linear combinations of $v$ and $w$?

Question

$\vec{u}=\left[\begin{matrix}2\\0\\0\end{matrix}\right]$, $\vec{v}=\left[\begin{matrix}0\\2\\2\end{matrix}\right]$, $\vec{w}=\left[\begin{matrix}2\\2\\3\end{matrix}\right]$

The linear combination of $\vec{u}, \vec{v}$ is $2\left[\begin{matrix}a\\b\\b\end{matrix}\right]$ where $a, b$ are some scalars.

The linear combination of $\vec{v}, \vec{w}$ is $\left[\begin{matrix}2d\\2c+2d\\2c+3d\end{matrix}\right]$ where $c, d$ are some scalars.

Is there an equation I can give for those vectors which are linear combinations of $\vec{u}$ and $\vec{v}$ and are also linear combinations of $\vec{v}$ and $\vec{w}$? If so, how?

Answer 1

Observe that the $u,v,w$ are linearly independent ( Using determinant ). Intersection of two subspaces is again a subspace.

What is the dimension of intersecting subspace?

We know, $dim(U + V) = dimU + dimV -dim(UV)$

Therefore the ans is $1(2+2-3)$

And we have the vector $v$ which belongs to the intersecting subspace.

Therefore the answer is linear span of $v$ i.e only vectors of the form $kv$ where k is real.

Answer 2

Suppose a vector $\vec{z}$ is a linear combination of both the pair $\vec{u}$ and $\vec{v}$ and the pair $\vec{v}$ and $\vec{w}$. In that case, we have the system of equations: $$a=2d$$ $$b=2c+2d$$ $$b=2c+3d$$ and this system has a nontrivial solution. Trivially, $a=d=0$, gives a solution which says that $\vec z=\lambda\vec v$ for any arbitrary scalar $\lambda$.

Attempting to solve the linear system results in concluding that only the trivial solution exists. If we subtract the second equation from the third, we get $b-b=(2c+3d)-(2c+2d)$, giving the conclusion $d=0$. However, as $a=2d$, this means that $a$ must also be $0$.