Choose such that the matrix whose rows are has rank . For example .
Fernando Revilla
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?
Choose such that the matrix whose rows are has rank . For example .
Fernando Revilla
It is not necessary, by a well known property, linearly independent vectors in a space of dimension always form a basis of .
Fernando Revilla
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?
Right.
Fernando Revilla
A more general (abstract) way of doing this. Since the original set is NOT a basis for R^4, there exist a vector in R^4 is not equal to a linear combination of and . Call such a vector and add it to the set: 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, in R^4 which cannot be written as a linear combination of , , and and so the new set is a set of 4 independent vectors in R^4 and so a basis.
Applying that concept to this problem, any linear combination of and 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
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.)