Let $A$ be a set of points that satisfies $\begin{bmatrix}a-b\\b-c\\c-a\\b \end{bmatrix}$, with $a,b,c \in \mathbb{R}$. Decide if $A$ is a vector space and if so, choose a set of vectors that span $A$. It seems a bit tedious to check for all the properties of a vector space, so I thought that it might be easier to check if $A$ is a subspace of another space, like $\mathbb{R}^4$.
- Id: $A$ contains the zero vector (e.g let $a=b=c=0$)
- Closure under addition: For $x,y,z\in\mathbb{R}$ we have $\begin{bmatrix}a-b\\b-c\\c-a\\b \end{bmatrix}$ + $\begin{bmatrix}z-y\\y-z\\z-x\\y \end{bmatrix}$=$\begin{bmatrix}(a+x)-(b+y)\\(b+y)-(c+z)\\(c+z)-(a+x)\\b+y \end{bmatrix}$.
- For scalar (real) multiplication I suppose it is quite straight forward.
I am having difficulties when it comes to finding a set of vectors of the given form that spans $A$. A hint or solution would be appreciated.
Best Answer
You have three parameters $a,b,c$ in the general vector. So you want to split up the general vector into a sum of three vectors, each of which involving only one parameter. What I mean is, write $$\begin{bmatrix}a-b\\b-c\\c-a\\b\end{bmatrix}=a\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}+b\begin{bmatrix}-1\\1\\0\\1\end{bmatrix}+c\begin{bmatrix}0\\-1\\1\\0\end{bmatrix}$$ so the first vector only involves $a$, the second only involves $b$, and the third only involves $c$. Then you can see that these three vectors (on their own, without the $a,b,c$) span the subspace.