Vector Space

A vector is a function from an index set I to the real numbers, x\colon I\to\boldsymbol{R}. Functions can be multiplied by a scalar, (ax)(i) = (xa)(i) = a(x(i)), and added, (x + y)(i) = x(i) + y(i), a\in\boldsymbol{R}, i,j\in I. This generalizes the usual definition of \boldsymbol{R}^n as the set of n-tuples of real numbers \{(x_1,\ldots,x_n)\mid x_j\in\boldsymbol{R}\} by letting I = \{1,\ldots,n\} and defining x by x(i) = x_i.

Exercise. Show a(x + y) = ax + ay for a\in\boldsymbol{R}, x,y\in\boldsymbol{R}^I.

Define the zero vector \boldsymbol{0}^I\colon I\to\boldsymbol{R} by \boldsymbol{0}^I(i) = 0 for i\in I. We write \boldsymbol{0} if I is understood.

Exercise. Show x + \boldsymbol{0} = x = \boldsymbol{0} + x for every vector x.

The zero vector is an additive identity for vector addition.

Exercise. Show if x + x = x then x = \boldsymbol{0}.

Hint: x(i) + x(i) = x(i) for all i\in I.

If A and B are sets then the set exponential B^A = \{f\colon A\to B\} is the set of all functions from A to B.

Exercise. If A and B are finite show \#B^A = \#B^{\#A}.

Hint: Where \#A is the number of elements in set A.

Define e_i\in\boldsymbol{R}^I, i\in I, by e_i^I(j) = 1 if i = j and e_i^I(j) = 0 if i\not=j, j\in I. The standard basis of the vector space \boldsymbol{R}^I is (e_i^I)_{i\in I}. We write e_i if I is understood.

Exercise. If I is finite show x = \sum_{i\in I}x(i)e_i for x\in\boldsymbol{R}^I.

A linear transformation is a function T|colon\boldsymbol{R}^I\to\boldsymbol{R}^J that preserves the vector space structure, T(ax) = a(Tx) and T(x + y) = Tx + Ty, a\in\boldsymbol{R}, x,y\in\boldsymbol{R}^I.

Exercise. If T is a linear transformation show T\boldsymbol{0}^I = \boldsymbol{0}^J

Hint: Use T(\boldsymbol{0} + \boldsymbol{0}) = T\boldsymbol{0}.

The set of linear transformations, \mathcal{L}(\boldsymbol{R}^I,\boldsymbol{R}^J), is also a vector space with scalar multiplication (aT)(x) = a(Tx) and addition (T + S)(x) = Tx + Sx, a\in\boldsymbol{R}, T,S\in\mathcal{L}(\boldsymbol{R}^I,\boldsymbol{R}^J). For i\in I and j\in J let t_{ij} = Te_i(j). We call (t_{ij})_{i\in I,j\in J} the matrix of T under the standard basis for \boldsymbol{R}^I and \boldsymbol{R}^J.

Exercise Show T(\sum_{i\in I} x_i e_i) = \sum_{j\in J}(\sum_{i\in I} t_{ij}x_i) e_j.

If T\in\mathcal{L}(\boldsymbol{R}^I,\boldsymbol{R}^J) and S\in\mathcal{L}(\boldsymbol{R}^J,\boldsymbol{R}^K) then the composition ST\in\mathcal{L}(\boldsymbol{R}^I,\boldsymbol{R}^K).

Exercise. Show the matrix of U = ST is u_{ik} = \sum_{j\in J}t_{ij}s_{jk}, $iI, k\in K.

Every T\in\mathcal{L}(\boldsymbol{R}^I,\boldsymbol{R}^J) corresponds to \overline{T}\in\boldsymbol{R}^{I\times J} by \overline{T}(i, j) = (Te_i)(j), i\in I, j\in J.

Exercise Show this correspondence is one-to-one and onto.

Hint: Show the standard basis of \boldsymbol{R}^{I\times J} is e_{ij}^{I\times J}(k, l) = e_i^I(k)e_j^J(l).