Linear Algebra: Determine whether S spans R^4

[LoA]

Homework Statement

Let S be a subset of $R^{4}$:
S = {$v_{1}$,$v_{2}$, $v_{3}$, $v_{4}$} = { [1,3,2,0] [-2,0,6,7] [0,6,10,7] [2,10,-3,1] }

Determine whether S spans $R^{4}$.

Homework Equations

span(S) = {V| V = a$v_{1}$+b$v_{2}$+c$v_{3}$+d$v_{4}$}

The Attempt at a Solution

When I row reduce the matrix A, whose rows are the vectors in S, I get a matrix whose rank = 3. So I conclude that S does NOT span $R^{4}$. Is this correct? If so, why? Let me explore a little further.

Let B = rref(A) =

1 0 0 $\frac{-221}{82}$
0 1 0 $\frac{59}{82}$
0 0 1 $\frac{11}{49}$
0 0 0 0

So rank(B) = 3 and span(S) = a[1,0,0$\frac{-221}{82}$] + b[0,1,0,$\frac{59}{82}$] +c[0,0,1,$\frac{11}{49}$] + d[0,0,0,0] for a,b,c,d $\in R$

i.e. span(S) = [a,b,c,D] ; D = $a\frac{-221}{82}$+$b\frac{59}{82}$ +$c\frac{11}{49}$ for a,b,c,d $\in R$

So why would it be the case that you cannot form every vector in $R^{4}$ as a linear combination of vectors in S? Is it because your choice of a,b,c determines D, so that you cannot have that same D for other, arbitrary choices of a,b,c (or different values of D for the same values of a,b,c)? How can I make this sharper? How should I think about this in terms of the rank of the matrix of vectors in S?

 

[Fredrik]

Homework Statement

Let S be a subset of $R^{4}$:
S = {$v_{1}$,$v_{2}$, $v_{3}$, $v_{4}$} = { [1,3,2,0] [-2,0,6,7] [0,6,10,7] [2,10,-3,1] }

Determine whether S spans $R^{4}$.

Homework Equations

span(S) = {V| V = a$v_{1}$+b$v_{2}$+c$v_{3}$+d$v_{4}$}

The Attempt at a Solution

When I row reduce the matrix A, whose rows are the vectors in S, I get a matrix whose rank = 3. So I conclude that S does NOT span $R^{4}$. Is this correct? If so, why? Let me explore a little further.

Let B = rref(A) =

1 0 0 $\frac{-221}{82}$
0 1 0 $\frac{59}{82}$
0 0 1 $\frac{11}{49}$
0 0 0 0

So rank(B) = 3 and span(S) = a[1,0,0$\frac{-221}{82}$] + b[0,1,0,$\frac{59}{82}$] +c[0,0,1,$\frac{11}{49}$] + d[0,0,0,0] for a,b,c,d $\in R$

i.e. span(S) = [a,b,c,D] ; D = $a\frac{-221}{82}$+$b\frac{59}{82}$ +$c\frac{11}{49}$ for a,b,c,d $\in R$

So why would it be the case that you cannot form every vector in $R^{4}$ as a linear combination of vectors in S? Is it because your choice of a,b,c determines D, so that you cannot have that same D for other, arbitrary choices of a,b,c (or different values of D for the same values of a,b,c)? How can I make this sharper? How should I think about this in terms of the rank of the matrix of vectors in S?

What you have found is that S is linearly dependent, and this is probably all you need to do to get full credit for the problem. The easiest way to do this was probably to use the theorem that says that a square matrix has determinant 0 if and only if its rows are linearly dependent. So if you form the matrix that has the components of those vectors as its rows, and prove that its determinant is 0, you will have proved that S is linearly dependent.

Since S is linearly dependent, one of its members is a linear combination of the other three. This implies that span S is equal to the span of the set whose members are those three vectors. So what you're asking is why a set with three members can't span $\mathbb R^4$. I don't see a super elegant way to prove this right now, but here's one way to prove it: Suppose that E={u,v,w} is a linearly independent set that spans $\mathbb R^4$. Let $x\in\mathbb R^4$ be arbitrary. Since E spans $\mathbb R^4$, {u,v,w,x} is linearly dependent. So the determinant of the matrix whose rows are the components of these four vectors is zero. Now you make a choice of x that contradicts that the value of the determinant is zero.