I have a set $$A = \{(1, -2, -1, 1), (2, 1, 3, 2), (1, 1, 3, 1), (0, 1, 2, 0)\}$$ and would like to know if it spans $\mathbb R^4$.
I built an augmented matrix to check if linear combinations of the vectors of A can construct all vectors in $\mathbb R^4$.
Since there is a free variable, the system is still consistent. My initial thought was that the set does span $\Bbb R^4$, since the free variable must just mean that there are an infinite number of linear combinations of the vectors in the set, which will construct the vectors of $\Bbb R^4$. However, upon further reading, it seems like that is not what it means. In fact, the bits and pieces I've researched seem to indicate that a free variable in this case means that the set is NOT a spanning set?
In previous spanning set problems, I have only had consistent systems with unique solutions. Can someone please solve the above problem and detail what the free variable means in this case? In the context of spanning sets, what is the difference between getting a consistent system with a unique solution, and a consistent system with an infinite number of solutions?
Thank you.
Best Answer
Hint:
the four vectors are linearly independent (so they span $\mathbb{R}^4$) only if the determinant of the matrix that has such vectors as columns (or rows) is not null. In this case the determinant is null.
This means that the system:
$$ \begin {cases} a+2b+c=0\\ -2a+b+c+d=0\\ -a+3b+3c+2d=0\\ a+2b+c=0 \end{cases} $$ has a not trivial solution.In other words, there is a solution of the system with a free variable. So there are four numbers, not all null, such that: $$ a\begin{bmatrix} 1\\-2\\-1\\1 \end{bmatrix}+ b\begin{bmatrix} 2\\1\\3\\2 \end{bmatrix}+ c\begin{bmatrix} 1\\1\\3\\1 \end{bmatrix}+ d\begin{bmatrix} 0\\1\\2\\0 \end{bmatrix}=0 $$