# Find a basis for the subspace $\mathbb{R}^3$ containing vectors

Let

$v_1 = \langle 1,0,-1\rangle$

$v_2 = \langle -2,7,2\rangle$

$v_3 = \langle 3,-7,-3\rangle$

I found that these are linearly dependent since I have a free variable upon reducing. However, the question asks to form a basis with those $3$ vectors. A basis can only be formed if all of the vectors are linearly independent.

How would I answer the following: "Find a basis for the subspace $\mathbb{R}^3$ containing $v_1$, $v_2$, $v_3$."?

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The subspace (I assume the subspace spanned by the given vectors, here) contains the vectors, not the basis. You can take the non-zero rows of your echelon form (assuming you work with the matrix whose rows are the given vectors) as the elements of the required basis. –  David Mitra Apr 26 '13 at 15:21
@DavidMitra Isn't $e_1,e_2,e_3$ also a basis than ? I think it is a strange question. –  Kasper Apr 26 '13 at 15:22
Note that the question is asking about the subspace $V$ of $\mathbb{R}^3$ which contains $v_1, v_2, v_3$, that is, $V = \operatorname{span}\{v_1, v_2, v_3\}$. The question is asking you to find a basis for $V$. –  Michael Albanese Apr 26 '13 at 15:24
That's my issue. How can I find a basis if the set is not linearly independent? Or am I confusing it all? –  Dimitri Apr 26 '13 at 15:29
Your question as stated and as implied by Kasper has a trivial answer. Perhaps you meant, as I suggest in my first comment, "the subspace spanned by the vectors"? –  David Mitra Apr 26 '13 at 15:29

I think you're overthinking this. These three vectors lie in a common plane. A plane requires only two basis vectors to be completely described. Therefore, you can pick any two vectors that are linear combinations of the given set of 3, and you're done.

In other words, choose two basis vectors $b_1, b_2$ and express them as linear combinations of $v_1, v_2, v_3$. You have freedom to choose what $b_1, b_2$ are, provided that they are linearly independent of one another.

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Thank you. I was overthinking it. It was a lot simpler than I thought. –  Dimitri Apr 30 '13 at 14:07

$v_3=v_1-v_2$ but $\{v_1,v_2\}$ are independ so they can be a basis for this subspace (that contain this three vectore)

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I'm just a student and get confused on the grammar but my comment is this: For a basis for R^3 you need 3 L.I. vectors to span R^3 not two. To get the third you can do the dot product with the first two to get a third L.I. vector letting one of the variables being anything. The point is this 3rd vector will be L.I. from the first two and not lie in the plane of the first two and therefore span R^3.

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Welcome to MSE! I realize you do not yet have enough reputation, but this is best as a comment rather than an answer. Regards –  Amzoti Jun 22 '13 at 5:39