# 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 $v_1$, $v_2$, $v_3$."?

• 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

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