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I'm not really sure how to extend a basis. I'm trying to do the following question.

Consider the subspace $ W = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 = -x_4, x_2 = x_3\}$ of $ \mathbb{R}^4$. Extend the basis $\{(0,2,2,0),(1,0,0,-1)\}$ of $W$ to a basis of $ \mathbb{R}^4$.

I know I need to add another two vectors for it to be a basis of $ \mathbb{R}^4$ but I'm not sure how to pick the vectors. In general, how do you expand a basis?

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Hint: Any $2$ additional vectors will do, as long as the resulting $4$ vectors form a linearly independent set. Many choices! I would go for a couple of very simple vectors, check for linear independence. Or check that you can express the standard basis vectors as linear combinations of your $4$ vectors.

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    $\begingroup$ I seem to be doing a lot of trial and error to find two additional vectors that are linearly independent of the current two. What are we actually doing when we expand a basis? The result is a basis for a new vector space, but do the restrictions in the "current" subspace have any relevance on the new basis? I tried the basis $\{(0,2,2,0), (1,0,0,-1), (1,0,0,0), (0,0,0,1)\}$ but that wasn't linear independent. How would you go about choosing the additional vectors? $\endgroup$ – user1520427 Aug 23 '12 at 5:48
  • $\begingroup$ Try $(1,0,0,0)$ and $(0,1,0,0)$ (plus the two you were given). $\endgroup$ – André Nicolas Aug 23 '12 at 5:49
  • $\begingroup$ Hmm I think I get it now. My one didn't work because the $(1,0,0,-1)$ was a linear combination of $\{(1,0,0,0), (0,0,0,1)\}$ correct? $\endgroup$ – user1520427 Aug 23 '12 at 5:51
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    $\begingroup$ @user1520427: Yes, that was the problem. Almost everything works, you had bad luck! $\endgroup$ – André Nicolas Aug 23 '12 at 5:55
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You might take a different 2-D subspace $X = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 = x_4, x_2 = -x_3\}$ which has a trivial intersection with $W$ and find a basis for it, for example $\{(0,2,-2,0),(1,0,0,1)\}$.

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  • $\begingroup$ and you can always find an example of such a space as the orthogonal complement, here the kernel of $\begin{pmatrix} 0 & 2 & 2 & 0 \\ 1 & 0 & 0 & -1 \end{pmatrix}$ $\endgroup$ – Cocopuffs Aug 23 '12 at 5:23
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Put those vectors into the rows of a matrix of A. By definition of row equivalence, the row reduced echelon form of A, call it R, has the same same row space as A. Now's it's easy to see which vectors are not in row(R). An easy solution for your extended vectors, for example, each should have a 1 at one of the free variables, and 0 at all the other variables.

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Write the given vectors with $e_1, e_2, e_3, e_4$, the standard basis of $\mathbb{R}^4$ in the columns of a matrix. By some row operations you will have four linearly independent vectors that will be basis.

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    $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Julian Kuelshammer Dec 18 '12 at 23:05
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    $\begingroup$ Could you specify what row operations you mean? I don't think the answer given is very clear (it may just be me!) $\endgroup$ – Tom Oldfield Dec 18 '12 at 23:09
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I suppose the easiest way to approach the problem is as follows. Consider the orthogonal complement of the subspace. A little trial and inspection shows that it is,

$ W^{\perp} = \{(y_1, y_2, y_3, y_4) \in \mathbb{R}^4 : y_1 = y_4, y_2 = - y_3\}$.

It is very easily to find a basis for this subspace as well. It is,

$ \beta=\{ (1,0,0,1), (0,1,-1,0) \}$.

Using the result that any vector space can be written as a direct sum of the a subspace and its orhogonal complement, one can derive the result that the union of the basis of a subspace and the basis of the orthogonal complement of its subspaces generates the vector space. You can proving it on your own.

Hence, take the union of $\beta$ and your basis to get the result.

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