Complete $\{(2,3,1),(1,4,3)\}$ to a basis of $\Bbb R^3$

Question

Having the following set: $C = \{(2,3,1),(1,4,3)\}$ I want to be able to generate $V = \mathbb{R}^3$

Since $V = \mathbb{R}^3$ has dimension $3$, $C$ needs $3$ elements. However, how do I find the vector I need to add to $C$ in order to be able to generate $V$?

Is sufficient to have 3 vectors that are Linearly Independent and then I can generate $V$? And how do I find it?

Thank you!

Answer 1

Take any vector $(a,b,c)$ which is not a linear combination of $(2,3,1)$ and $(1,4,3)$. For instance $(1,0,0)$ will do. Actually, any vector $(a,b,c)$ such that $b\neq a+c$ will do.

Answer 2

HINT

If you have 3 linearly independent vectors, they will span all of $\mathbb{R}^3$ (why?).

To get a third one, easiest way is to construct one perpendicular to them both, which is to take a cross product of the two you have

Answer 3

This method works in the general case and does not require using the "cross product":

Let $C=(v_1,v_2)$ as in the question. Choose any basis of $\mathbb{R}^3$. For simplicity we shall take $e_1,e_2,e_3$. Then look at

$(v_1,v_2,e_1)$, $(v_1,v_2,e_2)$ and $(v_1,v_2,e_3)$.

One of them is necessarily a linearly independent sequence of vectors (this can be checked by hand, for example you can take $e_1$ as in Jose' answer).

Indeed, if by contradiction all of them were linearly dependent then since $v_1,v_2$ are independent $e_1,e_2,e_3 \in \text{Span} \{v_1,v_2\}$. But this would mean that $\mathbb{R}^3 = \text{Span} \{v_1,v_2\}$ which is impossible.

Answer 4

Just take the cross product: $$(2,3,1) \times (1,4,3) = (5,-5,5)$$

Then $\{(2,3,1),(1,4,3),(5,-5,5)\}$ is linearly independent and hence a basis for $\mathbb{R}^3$.