# Find third vector to build a base of $\mathbb{R}^3$

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
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
@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
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
@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

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
@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
@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