## Mind-Warm-Ups Vector Spaces

Ultimate Space = S
si = arbitrary partition of S
Δsi = infinitesimal si, that is limit of si as it tends to zero.
pi = arbitrary point in S = Δsi

Problem 11: what is the dimension of the following vector spaces:
(i) V2(R)
(ii) V3(R)
Ans: (i) the dimension of V2(R) is 2
(ii) the dimension of V3(R) is 3.
Dimensions 1, 2 and 3 are the dimensions we perceive in the real world. However, mathematicians work with higher dimensions in advanced abstract analysis.

Problem 12: i = (1,0,0), j = (0,1,0), and k = (0,0,1) are basis for the vector space V3(R). Why?
Ans: i, j and k are basis for V3(R) because any vector v = (x, y, z)
can be written in the form xi + yj + zk.

Problem 13: define a family of points in a linear space X over the field F.
Ans: Let S be an index set and f a one-one mapping from S into X such that f(s) = xs, the points (xs) for all s in S, is a family of points in the linear space X.

Problem 14 : define a subfamily of the family (xs) of points in the linear space X over the field F.
Ans: Let S' be a subset of the index set S that generated the family of points (xs). If f, a one-one mapping from S into X is restricted to only the elements of S' such that f(s') = (xs'), then the points (xs') for all s' in S' is a subfamily of the points (xs) in the linear space X over the field F. The subfamily is finite if S' has a finite number of elements.

Problem 15 : how is the linear independence of the family (xs) of points in the linear space X determined?
Ans: Let αs be a set of scalars in the field F, and s an arbitrary member of the subset S' of the index set S. If for each subfamily (xs)s in S', ∑αsxs = 0 implies αs = 0 for all s in S', then the family (xs)s in S of points in X is said to be linearly independent. A family of points in X is linearly dependent if it is not linearly independent.

Problem 16: show that the basis i = (1,0,0), j = (0,1,0), and k = (0,0,1) in the space V3(R), are linearly independent.
Ans: let α1, α2, α3 be the scalars. Then ∑αsxs = α1(1,0,0) + α2(0,1,0) + α3(0,0,1) = (0, 0,0).
= (α1, α2, α3) = (0, 0, 0)
This implies α1 = α2 = α3 = 0. Therefore i, j, and k are linearly independent.
i, j, and k are called unit vectors of the vector space V3(R)
In general, the sum ∑αsxs is called a linear combination of the vectors and αs is called the coefficient of the vector xs.

Problem 17: if W is a subspace of the linear space X, explain the statement: W spans X.
Ans: W spans X if every element of X can be expressed as a linear combination of a finite number of elements of W. the vectors i, j, k of problem 16 span the vector space V3(R).

Problem 18: let xs be the range of a linearly independent family (xs)s in S of points in a linear space X (S is an index set)and let if xs span X. Therefore; xs is a Hamel basis. The number of elements in xs are infinite. True or False?
Ans: False.

Problem 19: the number of elements in a Hamel basis is called its cardinal number. This cardinal number is called the dimension of the space. If (ei, ..., en) is the Hamel basis of the space X, what is the dimension of X.
Ans: try it.

Problem 20: (ei, ..., en) is a Hamel basis for the space X; (ej, ..., em) is also a Hamel basis for the space X. If n = 7 what is (m + n) mod 14.
Ans: try it. hint: any two Hamel bases for a linear space has the same cardinal number.

Peter Oye Sagay