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Let $v_1 = (2, -1, -2)$ and $v_2 = (1, 2, -2)$

Find a vector $v_3$ such that $\{v_1,v_2, v_3\}$ is a basis of $\mathbb R^3$. Justify your answer.

Can someone please help me with this ? I have no clue how to start it or where to begin.

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closed as too localized by Andres Caicedo, anorton, Austin Mohr, 23rd, Micah Apr 26 '13 at 4:02

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Is this homework ? Do you know the definition of a basis ? –  Kasper Apr 26 '13 at 0:49
    
Not Homework, exam from last semester. And sadly do not know the definition of basis but looking it up. –  Reza M. Apr 26 '13 at 0:53
    
"have no clue how to start" ?? Seems like looking up the meaning of "basis" is a pretty obvious place to start. –  bubba Apr 26 '13 at 1:21
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I vote to close this question, as it provides no motivation, in line with meta.math.stackexchange.com/questions/9201/… –  Eric Naslund Apr 26 '13 at 1:33
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6 Answers 6

I would recommend first learning more about what a basis is. You could for example watch this video:

https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/subspace_basis/v/linear-algebra--basis-of-a-subspace

You may be able to solve this problem yourself after watching this video! If this video is too difficult, you could also start with a easier video from this list: https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces

I've watched all of those videos in that list, and they have been a great help for me in understanding the topics of linear algebra :)

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You just need to find a vector $v_3$ such that $v_1,v_2,v_3$ are linearly independent. This is the same as finding $v_3$ such that

$$\alpha_1v_1+\alpha_2v_2+\alpha_3v_3=0$$

only if $\alpha_1,\alpha _2,\alpha _3=0$.

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A simple technique to find a vector $v_3$ s.t. $v_1,v_2,v_3$ are linearly independant and hence we have a basis of $\mathbb{R}^3$ is to take for example $v_1+v_2$ or any linear combination and you and then perturb one component of this vector: $$v_1+v_2=(3,1,-4)$$ so take for example $$v_3=(2,1,-4)$$

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Looks good to me! –  amWhy Jul 25 at 12:34

The cross product of the two given vectors will work, as long as they are not parallel and neither of them is zero. So, given $v_1$ and $v_2$, the three vectors $v_1$, $v_2$, $v_1 \times v_2$ will form a basis.

The key requirement is that the third vector should not lie in the plane containing the first two. Or, saying it another way, the parallelipiped with the three vectors as edges should have some "fatness" (a non-zero volume). This is what gives the three vectors the power to generate (span) all of $\mathbb R^3$.

This is why my cross-product suggestion works, and why the "perturbation" idea given by Sami works, and why Andreas' "pick anything" strategy almost always works.

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See the link to basis:

...a basis is a set of linearly independent vectors that, in a linear combination, can represent every vector in a given vector space...In more general terms, a basis is a linearly independent spanning set.

Given a basis of a vector space, every element of the vector space can be expressed uniquely as a finite linear combination of basis vectors... Every vector space has a basis, and all bases of a vector space have the same number of elements, called the dimension of the vector space.

A basis for the vector space $\mathbb R^3$ requires that there be three vectors that are linearly independent

So your task is to find a vector $v_3$ such that $v_1, v_2, v_3$ are linearly independent.

Review your notes, and the text you use, to recall what it means for vectors to be linearly independent (and while you're at it, what it means for 3 such vectors to span $\mathbb R^3$).

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You don't need two conditions. If three vectors are linearly independent, then they automatically span $\mathbb R^3$. The converse is also true. –  bubba Apr 26 '13 at 1:13
    
@amWhy: nice write - up +1 –  Amzoti Apr 26 '13 at 1:33

Pick a vector at random; chances are it will serve as the required $v_3$. The reason is that the only vectors that don't work are those in the plane spanned by $v_1$ and $v_2$. With any reasonable randomization, a vector chosen at random in $3$-dimensional space is very unlikely to happen to lie on this one plane.

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