Write your four vectors as column vectors of a $ \ 5 \times 4 \ $ matrix and row reduce it:
$$ \left( \begin{array}{cc} 6 & 1 & 1 &7 \\4&0&4&1\\1&2&-9&0\\-1&3&-16&-1\\2&-4&22&3 \end{array} \right) \ \ \rightarrow \ \ \left( \begin{array}{cc} 0 & 0 & 0 &0 \\0&0&0&0\\0&0&0&1\\1&-3&16&1\\0&1&-5&\frac{1}{2} \end{array} \right) $$
[I used the fourth row here to work against the other rows; that doesn't matter particularly.]
What does the "zeroing-out" of two rows tell us? How can we use what non-zero rows remain to construct a basis for span(S) ? (Notice that these are five-dimensional vectors, so we are already starting out "short a coordinate variable", making it "free".)
EDIT -- Since the discussion has advanced further, we can say something about the basis of span(S). Taking the hint from Omnomnomnom or the above, the subspace spanned by your set of four vectors only has dimension 3. So we need to set up three linearly independent vectors, using the columns of the row-reduced matrix.
We could "reduce" those last two rows a bit more to obtain
$$ \rightarrow \ \ \left( \begin{array}{cc} 0 & 0 & 0 &0 \\0&0&0&0\\0&0&0&1\\1&0&1&0\\0&1&-5&0 \end{array} \right) \ \ . $$
With the matrix fully "reduced", we need to pick out three (five-dimensional) column vectors which are linearly independent. The third column is a linear combination of the first two, so we can toss that one out. A suitable basis for span(S) is then
$$ \left( \begin{array}{cc} 0 \\0\\1\\0\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\1\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\0\\1 \end{array} \right) \ \ . $$
Let $V$ be a vector space and $U$ a subspace of $V$. Then there is a subsace $W$ such that
$V= U \oplus W$. Define the linear mapping $P:V \to V$ as follows:
for $v \in V$ there are unique $u \in U$ and $w \in W$ such that $v=u+w$; put
$$Pv=w.$$
Then: $ker(P)=U$.
Can you get it from here ?
Best Answer
No. A subspace does not have to have the same dimension as the space it's from. In fact, $U$ actually has 4 basis vectors, since there is a linear relation between $a_{1}$ and $a_{3}$. To see why this would be true, think about R$^3$. This would be a subspace of R$^5$, which would be trivial to show, and we already know it has a dimension of 3.