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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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What are $\lambda_1,\lambda_2$, and $\lambda_3$, and why do you say that $\left(\begin{matrix} 3 & 3 & x \\ 2 & 2 & y \\ 0 & 1 & z \end{matrix}\right) \left(\begin{matrix} {\lambda}_1 \\ {\lambda}_2 \\ {\lambda}_3 \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right)$? –  littleO Nov 27 '12 at 11:59
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Yes, the vectors have to be linearly independent --- but that equation doesn't guarantee linear independence. What guarantees linear independence is insisting that the only solution of that equation be $\lambda_1=\lambda_2=\lambda_3=0$. –  Gerry Myerson Nov 27 '12 at 12:20
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@Flavius if there were a nonzero vector $\lambda = \begin{pmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \end{pmatrix}$ such that $\left(\begin{matrix} 3 & 3 & x \\ 2 & 2 & y \\ 0 & 1 & z \end{matrix}\right) \left(\begin{matrix} {\lambda}_1 \\ {\lambda}_2 \\ {\lambda}_3 \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right)$, this would mean that the columns of $\left(\begin{matrix} 3 & 3 & x \\ 2 & 2 & y \\ 0 & 1 & z \end{matrix}\right)$ were linearly dependent. –  littleO Nov 27 '12 at 12:21
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If you were given two linearly independent vectors in R^4 and wanted to extend them to a basis, you can do something similar: Get your two given vectors and two indeterminate vectors, stick them as the columns of a 4x4 matrix, reduce as far as possible with row/column operations, and make the final choices so that no row/column is zero. Here you could have divided column 1 by 3, cleared the top row, then used column two to clear the third column to get: $$\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -\frac{2}{3}x+y \end{matrix}\right).$$ –  Katie Dobbs Nov 27 '12 at 12:34
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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. –  littleO Nov 27 '12 at 12:48

4 Answers 4

up vote 2 down vote accepted

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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Maybe a link or some context would be helpful here? –  Simon Hayward Nov 27 '12 at 13:39
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@SimonHayward added the explanation. –  egens Nov 27 '12 at 14:27

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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Don't the vectors have to be linearly independent? –  Flavius Nov 27 '12 at 12:01
    
@Flavius Yes, and the cross product is a great way to ensure that they are. Have you seen the fact that the cross product of two vectors is normal to the plane spanned by those two vectors? –  Katie Dobbs Nov 27 '12 at 12:06
    
Is this true that the $-\frac{2}{3}x+y\neq 0$? –  Babak S. Nov 27 '12 at 12:08
    
@BabakSorouh Indeed that is a necessary and sufficient condition for (x,y,z) to complete the given basis. –  Katie Dobbs Nov 27 '12 at 12:09
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@Flavius: Bring it up as an another new question. :) –  Babak S. Nov 27 '12 at 12:19

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.

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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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How would I do it for $R^4$ if I would be missing the 4th vector? –  Flavius Nov 27 '12 at 12:08
    
A general rule that will work in ${\bf R}^n$ is, choose the last vector so the determinant is not zero. –  Gerry Myerson Nov 27 '12 at 12:21
    
It would be harder but you can probably do the same thing. –  Adam Rubinson Nov 27 '12 at 12:22
    
Give me a 4th vector if you don't believe me –  Adam Rubinson Nov 27 '12 at 12:26
    
What is certainly true in ${\bf R}^4$ is that at least one of the vectors $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$ is guaranteed to be a satisfactory choice as the 4th vector. You can try each one in turn, until you find one that works. –  Gerry Myerson Nov 27 '12 at 21:50

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