Problem:
Consider the set of vectors $S= \{a_1,a_2,a_3,a_4\}$ where
$a_1= (6,4,1,-1,2)$
$a_2 = (1,0,2,3,-4)$
$a_3= (1,4,-9,-16,22)$
$a_4= (7,1,0,-1,3)$
Find the dimension of the subspace $span(S)$?
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I know that dimension is the maximum number of linearly independent vectors in a subspace.
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So is the dimension in this case 4? Since there are 4 vectors?
Find a set of vectors in $S$ that forms basis of $span(S)$?
- How do I solve this?
Best Answer
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
$$ \left( \begin{array}{cc} 0 \\0\\1\\0\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\1\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\0\\1 \end{array} \right) \ \ . $$