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Question:

Given a spanning set of a subspace $U\subset \mathbb{R^3}$ where $U=\{(-1,-5,0),(-4,-21,-4),(1,10,20),(2,12,8)\}$ find a basis of $U$.

Solution

I first setup a matrix where each row was member of $U$ i.e.

$$\begin{bmatrix} -1 & -5 & 0 &|~~ 0\\ -4 & -21 & -4 &|~~ 0 \\ 1 & 10 & 20 &|~~ 0 \\ 2 & 12 & 8 &|~~ 0 \end{bmatrix}$$

Then proceed to get the matrix in echelon form using Jordan-Gaussian elimination. This gave $:$

$$\begin{bmatrix} 1 & 5 & 0 &|~~ 0\\ 0 & 1 & 4 &|~~ 0 \\ 0 & 0 & 0 &|~~ 0 \\ 0 & 0 & 0 &|~~ 0 \end{bmatrix}$$

Thus we look at the original matrix and take the corresponding non-zero rows, in this case it will be the first two elements of $U$. Hence a basis of $U$ is the following set, $$S=\{(-1,-5,0),(-4,-21,-4)\}$$

End solution.

Is this correct? Are there any more efficient ways of doing this as it takes a while and errors can slip in whilst doing JG elimination if you are not careful. Also does it make any difference how you form the original matrix or will it just give you a different basis? (as in could you swap the rows or even do a $3\times4$ matrix instead of a $4\times3$?)

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  • $\begingroup$ I think you set up your matrix wrong. Each $3$ dimensional vector should be a column, not a row. You should have three rows and four columns. I'm too rusty to remember if that makes a difference. $\endgroup$ – layman Mar 16 '15 at 21:00
  • $\begingroup$ I checked using an online calculator and it appears it does give different solutions. I'm not sure whether it would actually be a basis or not considering they are not unique. I'm sure there maybe an error in my calculation anyway. Ah well I'll set it up the other way next time just to be sure thanks. $\endgroup$ – Callum K Mar 16 '15 at 21:13
  • $\begingroup$ @user46944 It doesn't really matter if you put create the matrix by putting your vectors in the columns or the rows. The only difference is that if you put them in the rows (like OP did), you need to use row reduction (elementary row operations) and if you put them in the columns, you use column reduction (elementary column operations). I should mention though, even if you do the wrong type of reduction, you can still figure out a basis for the space... $\endgroup$ – user137731 Mar 17 '15 at 3:02
  • $\begingroup$ @Bye_World I must not remember my linear algebra very well. I could swear when I was learning this I used to put them as columns, and then use row reduction to get the answer. $\endgroup$ – layman Mar 17 '15 at 3:04
  • $\begingroup$ You can do it that way, too. But then you have to look at your orginal matrix to get your basis. Any columns with pivots in your row reduced matrix will correspond to which columns of the original matrix you get to keep. For instance, looking at OP's matrices, if he were looking for a linearly independent set of the columns of his first matrix, then his row reduced matrix would tell him that the first two columns span his column space. $\endgroup$ – user137731 Mar 17 '15 at 3:07
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Got the same. One can take $$\left\{ {{\rm{( - 1}}{\rm{, - 5}}{\rm{,0)}}{\rm{,( - 4}}{\rm{, - 21}}{\rm{,4)}}} \right\}$$

for a basis. It can't be more than two vectors. Because $U \subset {R^3}$.

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  • $\begingroup$ What? You mean it can't be more than 3 vectors if $U \subseteq \Bbb R^{3}$. $\endgroup$ – layman Mar 16 '15 at 22:41
  • $\begingroup$ The premission is $U \subset {R^3}$, that is U can't be the same as ${R^3}$, it must have less than three linear independent vectors. $\endgroup$ – Frieder Mar 16 '15 at 22:51
  • $\begingroup$ Only if you assume $\subset$ is the symbol for proper subset. I've seen plenty of cases where people don't assume this, so we can't say for sure $U$ cannot equal $\Bbb R^{3}$ when we first look at the problem. $\endgroup$ – layman Mar 16 '15 at 22:53
  • $\begingroup$ Yes! Assumption is, that we are looking for proper subset. $\endgroup$ – Frieder Mar 16 '15 at 23:00

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