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Find a basis for $$\mathbb{R}^4$$ such that the set contains the vectors:

$$\{ \langle 1, 2, -1, 3\rangle, \langle 0, 2, 1, 1 \rangle \}$$

I know that I need to find 2 vectors $v_1, v_2$ such that

$$\{ \langle 1, 2, -1, 3\rangle, \langle 0, 2, 1, 1 \rangle, v_1, v_2 \}$$

Is a linearly independent set. But what is a strategy for doing that? This seems variable and it doesnt seem straight forward?

Marodian
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  • Does your textbook discuss anything about extension of a linearly independent set to a basis in some vector space? Here (http://math.stackexchange.com/q/465870/9464) is a general way to do it. –  Mar 11 '17 at 16:32
  • Can you Gram-Schmidt this ish? – Sean Roberson Mar 11 '17 at 16:33
  • Do you know how to find the null space of a matrix? – amd Mar 11 '17 at 16:34

5 Answers5

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One systematic approach is as follows: begin by choosing any basis $\{e_1,e_2,e_3,e_4\}$; I will take the standard basis with $e_1= (1,0,0,0)$ and so on. We then row-reduce the matrix $$ \pmatrix{v_1&v_2&e_1&e_2&e_3&e_4} $$ In this case, you should end up with something of the form $$ \pmatrix{1&0&0&0&*&*\\ 0&1&0&0&*&*\\ 0&0&1&0&*&*\\ 0&0&0&1&*&*} $$ which is to say that the first four columns are "pivot columns". Correspondingly, we may conclude that the first four vectors $\{v_1,v_2,e_1,e_2\}$ form a basis of $\Bbb R^4$. Whichever basis of $e_i$ you happen to choose, the vectors corresponding to the eventual pivot columns will form a basis.

Ben Grossmann
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A convenient choice if you’re working in $\mathbb R^n$ or similar vector spaces is to use the basis of the orthogonal complement of your subspace. Since the row space of a matrix is the orthogonal complement of its null space, you can compute this by finding the null space of the matrix that has the given spanning vectors as rows.

In your case, we would find a basis for the null space of $$\pmatrix{1&2&-1&3\\0&2&1&1}$$ which can be done via row-reduction: $$\pmatrix{1&2&-1&3\\0&2&1&1}\to\pmatrix{1&0&-2&2\\0&1&\frac12&\frac12}$$ from which we can read the basis $(-2,\frac12,-1,0)^T$ and $(2,\frac12,0,-1)^T$.

amd
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  • I think you've done the row reduction slightly incorrectly. – Amad27 Mar 12 '17 at 06:37
  • Great answer still, so the basis of $$R^4$$ will be the original set plus these vectors transposed extended? – Amad27 Mar 12 '17 at 06:38
  • @Amad27 Oops. Good catch. Had it right on paper, but didn’t copy the first row correctly. Didn’t check that the product was zero as I usually do. And yes, the full basis would be the one you started with plus the null space basis. As well, the rows with pivots give you a basis for the original subspace, which is handy in case you had a linearly-dependent spanning set instead of a proper basis for the space. Omnomnomnom’s answer also sorts that case out. – amd Mar 12 '17 at 17:07
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I claim that $\{<1,2,-1,3>,<0,2,1,1>,<0,0,1,0>,<0,0,0,1>\}$ is linearly independent. To see this, note that $\det\begin{pmatrix} 1 & 2 & -1 & 3 \\ 0 & 2 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = 2$, so the matrix is non-singular, thus the rows are linearly independent.

Lukas Heger
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  • This doesn’t really answer the question. The OP wants to know how to go about finding a basis extension. Your vectors appear out of thin air. – amd Mar 11 '17 at 16:35
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General way is to take a pattern to get first three values, change the pattern for fourth value. Adding vectors gives ${1,4,0,4}$, break the pattern for last entry, making ${1,4,0,5}$ a basis. Same method for next basis but this time use all three vectors.

jnyan
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I am not sure if this answer is essentially the same as Omnomnomnom's, but here goes.

There is a theorem stating that given a linearly independent set $L$ and a generating set $S$ for the vector space, and $L\subseteq S$, there is a basis $B$ such that $L\subseteq B\subseteq S$.

When the vector space is finite-dimensional, there is an algorithm applying this theorem. In your case, the set $L$ is $\{ \langle 1, 2, -1, 3\rangle, \langle 0, 2, 1, 1 \rangle \}$. While for $S$, you can choose any finite generating set $S'$ you like, and choose $S:=S'\cup L$, which would be a finite generating set and a superset of $L$.

You then remove vectors from $S$ one by one to get a smaller and smaller generating set (and do not remove vectors from $L$) until you end up with a linearly independent generating set, which is a basis containing $L$.

This method should work well for other finite-dimensional vector spaces.

edm
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  • This is ostensibly the dual version to my approach. What I've effectively done is as follows: begin with a linearly independent set ${v_1,v_2}$. Add vectors from a spanning set. At each step, if ${v_1,\dots,v_{k-1},v_k}$ is linearly dependent, throw out the vector $v_k$ and add in the next vector. If ${v_1,\dots,v_{k-1},v_k}$ is linearly independent, keep $v_k$ and add in the next vector. – Ben Grossmann Mar 11 '17 at 17:39
  • That being said, my approach can be nicely framed in terms of row-reduction, which is an advantage for those coming in from a more computational side. – Ben Grossmann Mar 11 '17 at 17:41