Completing to a basis?

Factual Questions
1

Hi, I’m a CS undergraduate and we are currently studying linear algebra. One of our homework questions is regarding linear independence and completing to a basis. I’ve already proven that the three vectors in question are linearly independent, as requested, but now I’m stuck on how to complete to a basis. I’ve put the vectors into a (5 wide x 3 tall) matrix and put it into echelon form, however my course notes now state that I must add elementary vectors to the columns without leading terms, which in this case is row three and row five. However, my course notes just state this, and I’m confused, seeing as I thought the elementary vectors e3 and e5 were (0, 0, 1, 0, 0) and (0, 0, 0, 0, 1) (working in R5) repsectively? So now how do I add e3 and e5 to the third and five columns when they’re only 3 elements in height?

I have searched Google with the phrase “Completing to a basis” but that only brought up 3 sites (from universities asking homework questions) and I’ve also looked on Dr. Math, but that only comments on finding the basis for a vector space, not on completing to one. Mathworld at Wolfram also has no results found when I do a search for it.

I think my problem is with having a misunderstanding of the elementary vectors, or something like that. Could someone please clear this up for me? If it’s possible that completing to a basis is known as something else, could you tell me the alternate name?

Thanks.

2

If the three vectors that you were given to start with were each of length five (i.e. they each have five elements) then I assume you’re being asked to complete to a basis for the entire space R5 (i.e. the space of numbers in five dimensions). In this case any such basis must be five vectors of length five, or in other words if you are viewing them as row vectors of a matrix then it would have to be a 5 by 5 matrix. There is no way that the rows of a 5 by 3 matrix can span a space of five dimensions.

3

“For each of the subsets of R5 show that it is linearly independent and complete it to a basis by adding to it any needed elementary vectors”

(The following vectors are from a different question, as I don’t want to post the question I’m answering):

  1. {(1,1,1,1,1), (1, 1, 1, 1, -1), (1, 1, 1, -1, -1)}.

There’s four parts to the question, although we only have to do the first one. Each part only specifies three vectors, like above.

4

I’ll take a stab with explaining this, but someone who knows better will likely come along soon and correct me. If you’re vectors are linearly independent, then what they span is a 3-dimensional subspace of R5. You need to insert two vectors (linearly independent of the other three) that span the additional two dimensions in order to make a basis for R5. As ITR says, this will give you a 5x5 matrix in your case. You could never have a basis for R5 with a matrix that is Nx5, where N<5 (you could have a generalization of a basis when N>5, but then the elements are no longer linearly independent).

So, when you say you put it in echelon form, I assume you have a matrix that looks something like:

[0 0 x x x
0 0 0 x x
0 0 0 0 x]

where the x just marks entries that are non-zero. Is that right? If this is the case, can you see that you are only spanning dimensions 3-5 in R5, and dimensions 1-2 are not covered at all? Think about it by multiplying your matrix on the right by a vector v. When v=[1 0 0 0 0] or v=[0 1 0 0 0], the result will be all zeros. So, any vector with energy in dimensions 1-2 will get lost here, because the elements you have don’t span those dimensions. So, to correct the problem, insert the two necessary elementary vectors (e1 and e2, just like you defined them) to make the collection span all of R5.

5

Cheers, I’ve worked it out now. What I don’t understand is why there’s no websites that mention this procedure. Is it known under a different name or something else?

6

I’ve never heard the term “complete to a basis”. I’ll check books when I get home for what it’s called.

7

What you’re essentially doing when using crozell’s procedure is finding the nullspace of the 3x5 matrix whose rows are the given (linearly independent) vectors in R[sup]5[/sup]. Since the nullspace of that matrix is the orthogonal complement of its rowspace, you end up finding two more vectors that, together with the independent vectors given in the problem, span all of R[sup]5[/sup]. One might refer to this algorithm as an orthogonal decomposition of R[sup]5[/sup].

8

Odd, I’ve found it more or less common. In case anybody missed this class, usually it goes something like this:

Thm: Given a finite-dimensional vector space V (over R is usually assumed at this level), any collection of linearly independant vectors may be completed to a basis of V. That is, given {u[sub]1[/sub], …, u[sub]m[/sub]} one may find a basis {v[sub]1[/sub], …, v[sub]n[/sub]} of V with v[sub]i[/sub] = u[sub]i[/sub] for i between 1 and m.

The more advanced students will please note that this (a) essentially requires the assumption of finite-dimensionality, (b) essentially requires that we’re talking about vector spaces (opp. modules, where not all short exact sequences split.