4.1 Vector Spaces and Subspaces 4.2 Null Spaces, Column Spaces, and Linear Transformations 4.3 Linearly Independent Sets; Bases 4.4 Coordinate systems 4 Vector Spaces

Definition Let H be a subspace of a vector space V. An indexed set of vectors in V is a basis for H if i) is a linearly independent set, and ii) the subspace spanned by coincides with H ; i.e. REVIEW

The Spanning Set Theorem Let be a set in V, and let. a.If one of the vectors in S, say, is a linear combination of the remaining vectors in S, then the set formed from S by removing still spans H. b. If, some subset of S is a basis for H. REVIEW

Theorem The pivot columns of a matrix A form a basis for Col A. REVIEW

4.4 Coordinate Systems

Why is it useful to specify a basis for a vector space? One reason is that it imposes a “coordinate system” on the vector space. In this section we’ll see that if the basis contains n vectors, then the coordinate system will make the vector space act like R n.

Theorem: Unique Representation Theorem Suppose is a basis for V and is in V. Then there exists a unique set of scalars such that.

Definition: Suppose is a basis for V and is in V. The coordinates of relative to the basis (the - coordinates of ) are the weights such that. If are the - coordinates of, then the vector in is the coordinate vector of relative to, or the - coordinate vector of.

Example: 1. Consider a basis for, where Find an x in such that. 2. For, find where is the standard basis for.

on standard basison

The Coordinate Mapping Theorem Let be a basis for a vector space V. Then the coordinate mapping is a one-to-one and onto linear transformation from V onto.

Example: For and, find. For, let. Then is equivalent to. : the change-of-coordinates matrix from β to the standard basis

Example: Let Determine if x is in H, and if it is, find the coordinate vector of x relative to.

Application to Discrete Math Let  = {C(t,0), C(t,1), C(t,2)} be a basis for P 2, so we can write each of the standard basis elements as follows: C(t,0) = 1(1) + 0t + 0t 2 C(t,1) = 0(1) + 1t + 0t 2 C(t,2) = 0(1) – ½ t + ½ t 2 This means that following matrix converts polynomials in the “combinatorics basis” into polynomials in the standard basis:

Application to Discrete Math Recall that  = {C(t,0), C(t,1), C(t,2)} is a basis for P 2. The following matrix converts polynomials in the “combinatorics basis” to polynomials in the standard basis: Therefore, the following matrix converts polynomials in the the standard basis to polynomials in “combinatorics basis”:

The polynomial p(t) = 3 + 5t – 7t 2 has the following coordinate vector in the standard basis S = {1, t, t 2 }: We now want to find the coordinate vector of p(t) in the “combinatorics basis”  = {C(t,0), C(t,1), C(t,2)}: That is, p(t) = 3 C(t,0) – 2 C(t,1) – 14 C(t,2) Application to Discrete Math

Find a formula in terms of n for the following sum Solution. Using the form we found on the previous slide

Aside: Why are these true?

Application to Discrete Math Find the matrix M that converts polynomials in the “combinatorics basis” into polynomials in the standard basis for P 3 : Use this matrix to find a version of the following expression in terms of the standard basis:

Application to Discrete Math Find a formula in terms of n for the following sum Solution. Continued…..

Final Steps… We can finish by multiplying by a matrix that converts vectors written in {1,(t+1),(t+1) 2,(t+1) 3 } coordinates to a vector written in terms of the standard basis. Find such a matrix and multiply it by the answer on the previous slide to get a final answer of the form