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

For the following vectors $v_1 = \left(\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}\right)$ and $v_2 = \left(\begin{matrix} 2 \\ 2 \\ 5 \end{matrix}\right)$, find a third vector $v_3 = \left(\begin{matrix} x \\ y \\ z \end{matrix}\right)$ which together build a base for $\mathbb{R}^3$.*

My solution

We know that the vectors must be linearly independent, so

$$\left(\begin{matrix} 1 & 2 & x \\ 1 & 2 & y \\ 2 & 5 & z \end{matrix}\right) \left(\begin{matrix} {\lambda}_1 \\ {\lambda}_2 \\ {\lambda}_3 \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right) $$

for ${\lambda}_1 = {\lambda}_2 = {\lambda}_3 = 0$

The gauss reduction gives

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

So the solution would be any vector $v_3$ for which the following hold:

$$ y \neq x \\ z \neq 2x \\ x \neq 0 $$

for example $$ \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right) $$

Is this a valid solution?

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marked as duplicate by user1551, Micah, rschwieb, draks ..., Pedro Tamaroff Nov 27 '12 at 14:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You only need $y\neq x$ as a condition for the third vector be LI with others. – fmoura2005 Nov 27 '12 at 12:54
up vote 1 down vote accepted

That's a good approach and the answer is correct. when you think about what you would do with the Gauss reduction, it is only required that the bottom right entry $y-z$ is non-zero for the third vector to be linearly independent from the other two.

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So if I would get the bottom right $0$ after the reduction, then the first two vectors are already linearly dependent and there's no way to find $v_3$ such that the condition is right? – Flavius Nov 27 '12 at 12:59
Correct. You can only increase the dimension of the column space by 1 with each additional vector. – Hans Engler Nov 27 '12 at 13:07

After the reduction, the determinant of the matrix is $y-x$, so you get a basis if and only if $x \neq y$.

An alternative way to describe this, is that the first two vectors are not colinear, so they span a plane in $\Bbb R^3$. It is also obvious that they span the plane of equation $x=y$ because both vectors satisfy that equation. So in order to get a basis, you only need to add a vector which is not in that plane, that means a vector $(x,y,z)$ with $x \neq y$.

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