Extending a set to a basis more than one dimension away

Question

Say for instance I have the following set of vectors: $ \{(1, 2, 3, 4),(4, 3, 2, 1)\}$

If I want to extend this set to a basis of $\mathbb{R}^4$, would it be possible? I appreciate the help!

Answer 1

Any linearly independent set can be extended to a basis. Thus since $(1,2,3,4),(4,3,2,1)$ are easily seen to be linearly independent, then such an extension is certainly possible.

In general, it's pretty "hard" to make a "bad" choice of a vector to add (there is a sense where almost any vectors will work). Thus just add two random vectors and see if they work, e.g. $(1,0,0,0), (0,1,0,0)$. You can check this does actually form a basis.

Answer 2

Yes, it is possible. We will use the GS process to construct a basis which will be orthogonal as well.

It is easy to check that the vectors provided are linearly independent. Now, we will add two more vectors to the set, so that we have $4$ vectors to start with. In the GS process, we don't require the set to be linearly independent. We just put the vectors that are not an obvious multiple of the given vectors.

Let $(0,1,0,0)$ and $(0,0,1,0)$ be two other vectors.

Let $v_1=(0,1,0,0)$
Now, $v_2=(0,0,1,0)-\frac{<(0,0,1,0),v_1>}{||v_1||^2}v_1=(0,0,1,0)$
$v_3=(1,2,3,4)-\frac{<(1,2,3,4),v_1>}{||v_1||^2}v_1-\frac{<(1,2,3,4),v_2>}{||v_2||^2}v_2=(1,0,0,4)$
$v_4=(4,3,2,1)-\frac{<(4,3,2,1),v_1>}{||v_1||^2}v_1-\frac{<(4,3,2,1),v_2>}{||v_2||^2}v_2-\frac{<(4,3,2,1),v_3>}{||v_3||^2}v_3=(\frac{60}{17},0,0,-\frac{15}{17})$

So, the basis is $\{(0,1,0,0),(0,0,1,0),(1,0,0,4),(\frac{60}{17},0,0,-\frac{15}{17})\}$

Here are the few things for you to check quickly:
$1$ Check whether $\{(1,2,3,4),(4,3,2,1)\}$ is a Linearly independent set.
$2$ When can GS Process be applied?
$3$ Why should be start with $4$ vectors?