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For the following vectors $v_1 = (3,2,0)$ and $v_2 = (3,2,1)$, find a third vector $v_3 = (x,y,z)$ which together build a base for $\mathbb{R}^3$.

My thoughts:

So the following must hold:

$\left(\begin{matrix} 3 & 3 & x \\ 2 & 2 & y \\ 0 & 1 & z \end{matrix}\right) \left(\begin{matrix} {\lambda}_1 \\ {\lambda}_2 \\ {\lambda}_3 \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right) $

The gauss reduction gives

$ \left(\begin{matrix} 3 & 3 & x \\ 0 & 1 & z \\ 0 & 0 & -\frac{2}{3}x+y \end{matrix}\right) $

(but here I'm not sure if I'm allowed to swap the $y$ and $z$ axes)

For ${\lambda}_1 = {\lambda}_2 = {\lambda}_3 = 0$, this gives me

$ x = 0 \\ y = 0 \\ z = 0 $

Is this third vector $v_3$ building a base of $\mathbb{R}^3$ together with the other two vectors? If not, where are my mistakes?

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    @Flavius yes, that's right. You can choose the third column so that when you do row reduction, you don't end up with a row of zeros.2012-11-27

4 Answers 4

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Well (3,2,0) and (3,2,1) give you (0,0,1). So for your third one (0,1,0) would work.

Then (1,0,0) = 1/3 [(3,2,0) - 2(0,1,0)]

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    Yes....exactly.2012-11-28
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The big mistake is at the very beginning --- there is no reason at all why you should want that equation to hold.

There are infinitely many correct choices for $v_3$. One simple one is the cross product of $v_1$ and $v_2$ (warning --- this choice won't be available in other vector spaces).

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    Is it ok if I accept this answer and ask another question with a similar exercise to see if I've understood it right?2012-11-27
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There is more general solution, that assumes finding normalized basis of given linear subspace and then complement it to full basis by solving several homogeneous systems.

Basis normalisation. Suppose having $m$ linear independent vectors $\tilde{v}_1..\tilde{v}_m$ in $R^n$. Linear independence says that they form a basis in some linear subspace of $R^n$. To normalize this basis you should do the following:

  1. Take the first vector $\tilde{v}_1$ and normalize it $v_1 = \frac{\tilde{v}_1}{||\tilde{v}_1||}.$
  2. Take the second vector and substract its projection on the first vector from it $\bar{v}_2 = \tilde{v}_2 - (\tilde{v}_2 \cdot v_1) {v}_1,$ there $(\tilde{v}_2 \cdot v_1)$ is scalar product and equals to the length of projection, cosider $||v_1||=1$. Normalize $v_2 = \frac{\bar{v}_2}{||\bar{v}_2||}.$
  3. Take the $i=3..m$ vector $\tilde{v}_i$. Substract it projections on the all previously generated vectors of normal basis from it $\bar{v}_i = \tilde{v}_i - \sum_{j=1}^{i-1}(\tilde{v}_i \cdot v_j) {v}_j,$ and normalize it $v_i = \frac{\bar{v}_i}{||\bar{v}_i||}.$

Vectors $v_1..v_m$ will form new normalized basis. All their lengths are equal to 1 and they are normal to each other.

Homogeneous systems. To get the $(m+1)$'th basis vector $v_{m+1}$ the next homogeneous system of scalar productions must be solved $\begin{cases} v_1 \cdot v_{m+1} = 0 \\ v_2 \cdot v_{m+1} = 0 \\ ... \\ v_m \cdot v_{m+1} = 0 \end{cases}$

The solution of this system will be subspace, that is normal to given. One of its vectors should be taken as $v_{m+1}$.

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    @SimonHayward added the explanation.2012-11-27
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But we’re talking about vector spaces over $\mathbb R$ here. If the dimension of the vector space is $n$, then any set of fewer than $n$ vectors spans a lower-dimensional subspace, whose complement is open and dense in the whole. You should think of this as telling you that one more vector has almost no chance of being a wrong choice. So in the case at hand, any randomly-chosen third vector should complete a basis. Like $(5,-11,17/3)$, for example.