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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 $\left(\begin{matrix} 3 & 3 & x \\ 2 & 2 & y \\ 0 & 1 & z \end{matrix}\right) \left(\begin{matrix} {\lambda}_1 \\ {\lambda}_2 \\ {\lambda}_3 \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right)$?2012-11-27
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    @littleO Don't the vectors have to be linearly independent?2012-11-27
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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 $\lambda_1=\lambda_2=\lambda_3=0$.2012-11-27
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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*.2012-11-27
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    @littleO So a good solution to this one would be $(2, 1, 3)$ because $1\neq \frac{2}{3}\cdot 2$?2012-11-27
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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: $$\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -\frac{2}{3}x+y \end{matrix}\right).$$2012-11-27
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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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    How would I do it for $R^4$ if I would be missing the 4th vector?2012-11-27
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    A general rule that will work in ${\bf R}^n$ is, choose the last vector so the determinant is not zero.2012-11-27
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    It would be harder but you can probably do the same thing.2012-11-27
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    Give me a 4th vector if you don't believe me2012-11-27
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    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.2012-11-27
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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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    Don't the vectors have to be linearly independent?2012-11-27
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    @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?2012-11-27
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    Is this true that the $-\frac{2}{3}x+y\neq 0$?2012-11-27
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    @BabakSorouh Indeed that is a necessary and sufficient condition for (x,y,z) to complete the given basis.2012-11-27
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    @KatieDobbs Yes I've seen that, but how about higher dimensions of $\mathbb{R}$?2012-11-27
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    @KatieDobbs: So the OP could choose any triple $(x,y,z)$ in which $z$ is free but $-\frac{2}{3}x+y$should not be zero.2012-11-27
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    @Flavius: Bring it up as an another new question. :)2012-11-27
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    @BabakSorouh Indeed. We can look back at our vectors and see why this must hold as well: The span of the given two vectors contains $(0,0,1)$ so we can subtract off $z$ times that our third vector to see what is important is just $x$ and $y.$ The $x$ and $y$ of the first two is just $3$ and $2$ so to avoid linear dependence with them, we just need our $x$ and $y$ to not be a scalar multiple of those.2012-11-27
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    @KatieDobbs: Exactly! Thanks Katie. I voted up for this answer +1.2012-11-27
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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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    Maybe a link or some context would be helpful here?2012-11-27
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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.