# Extending a linearly independent set

• Feb 11th 2011, 12:21 AM
PatrickM
Extending a linearly independent set
Let u1 = (2; 1; 1; 1) and u2 = (4; 2; 2;-1).How can I extend the linearly independent set u1 and u2 to obtain a basis of R^4?.
I know that u1 and u2 are linearly independent since both vectors are non-zero and none is a multiple of the other ,but what shoud I do next?
• Feb 11th 2011, 12:29 AM
FernandoRevilla
Choose $u_3,u_4$ such that the matrix whose rows are $u_1,u_2,u_3,u_4$ has rank $4$ . For example $u_3=(1,0,0,0),u_4=(0,1,0,0)$ .

Fernando Revilla
• Feb 11th 2011, 12:35 AM
PatrickM
And since they are linearly independent I calculate to see if it spans R^4?
• Feb 11th 2011, 12:45 AM
FernandoRevilla
Quote:

Originally Posted by PatrickM
And since they are linearly independent I calculate to see if it spans R^4?

It is not necessary, by a well known property, $n$ linearly independent vectors in a space $V$ of dimension $n$ always form a basis of $V$ .

Fernando Revilla
• Feb 11th 2011, 12:50 AM
PatrickM
so if a linearly independent subset of V has most n elements; if it has n elements then it is a
basis of V since dim V=n?.....so all I have to do to solve this problem is to find 2 additional vectors(since R^4) and make sure they all 4 of the are linearly independent?
• Feb 11th 2011, 01:10 AM
FernandoRevilla
• Feb 11th 2011, 04:24 AM
HallsofIvy
A more general (abstract) way of doing this. Since the original set $\{u_1, u_2\}$ is NOT a basis for R^4, there exist a vector in R^4 is not equal to a linear combination of $u_1$ and $u_2$. Call such a vector $u_3$ and add it to the set: $\{u_1, u_2, u_3\}$ is now a set of independent vectors. If it spans R^4, we are done (obviously it doesn't since it has only three vectors and R^4 has dimension 4). If it does not span R^4 then there exist a vector, $u_4$ in R^4 which cannot be written as a linear combination of $u_1$, $u_2$, and $u_3$ and so the new set $\{u_1, u_2, u_3, u_4\}$ is a set of 4 independent vectors in R^4 and so a basis.

Applying that concept to this problem, any linear combination of $u_1$ and $u_2$ is of the form a(2; 1; 1; 1)+ b(4; 2; 2;-1)= (2a+ 4b,; a+ 2b; a+ 2b; a- b). Notice that the second and third components are equal. (0; 1; -1; 0) does NOT have that property and so cannot be written as a linear combination of those two vectors. Take $u_1= (0; 1; -1; 0)$

A linear combination of those three vectors is of the form a(2; 1; 1; 1)+ b(4; 2; 2;-1)+ c(0; 1; -1; 0)= (2a+ 4b; a+ 2b+ c; a+ 2b- c; a- b). Now look for a relation satisfied by those components. Since we just added the "c", it might be simplest to start by getting rid of it! If we write that as (x; y; z; u)= (2a+ 4b; a+ 2b+ c; a+ 2b- c; a- b), then, no matter what a, b, and c are separately, y+ z= x for all (x; y; z; u) in the span. And (1; 0; 0; 0) obviously does NOT satisfy that and so is not in the span. Adding that gives 4 independent vectors and so is a basis for R^4.

(Notice that there are many, many possible choices and an infinite number of correct solutions.)