Question M20 Page 403

The set B = {v1, v2, v3, v4} is a basis of the vector space P3, polynomials with degree 3 or less. Therefore ρB is a linear transformation, according to Theorem VRLT. Find a “formula” for ρB. In other words, find an expression for ρB(a + bx + cx^2 + dx^3).

v1 = 1 - 5x - 22x^2 + 3x^3

v2 = -2 + 11x + 49x^2 - 8x^3

v3 = -1 + 7x + 33x^2 - 8x^3

v4 = -1 + 4x + 16x^2 + x^3

Step 1

Determine the meaning of the question. In this question, ρB is a linear transformation going from P3 to \(C^4\). The basis of \(C^4\) can be defined as {1, x, \(x^2\), \(x^3\)}.

Step 2

Determine the vector representations of the elements of \(C^4\). This is done by solving the systems of equations from the set B.

I will show the set up for solving each system of equations and use R to come to each of the results.

C = {1, x, \(x^2\), \(x^3\)}, which can also be represented in vectors as 1{1,0,0,0}, x{0,1,0,0,}, \(x^2\){0,0,1,0}, and \(x^3\){0,0,0,1}.

The matrix used on the left side of the following equations is based on the original vectors from the question.

For 1{1,0,0,0}:

\(\begin{bmatrix} 1 & -2 & -1 & -1 \\ -5 & 11 & 7 & 4 \\ -22 & 49 & 33 & 16 \\ 3 & -8 & -8 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\)

Row reduce to solve the system of equations:

\(\begin{bmatrix} 1 & -2 & -1 & -1 & 1\\ -5 & 11 & 7 & 4 & 0\\ -22 & 49 & 33 & 16 & 0 \\ 3 & -8 & -8 & 1 & 0 \end{bmatrix}\)

library(pracma)
## Warning: package 'pracma' was built under R version 4.3.2
A <- matrix(c(1, -2, -1, -1, 1, -5, 11, 7, 4, 0, -22, 49, 33, 16, 0, 3, -8, -8, 1, 0), 4, 5, byrow = TRUE)
part1 <- rref(A)[,5]

For x{0,1,0,0}:

\(\begin{bmatrix} 1 & -2 & -1 & -1 \\ -5 & 11 & 7 & 4 \\ -22 & 49 & 33 & 16 \\ 3 & -8 & -8 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}\)

\(\begin{bmatrix} 1 & -2 & -1 & -1 & 0\\ -5 & 11 & 7 & 4 & 1 \\ -22 & 49 & 33 & 16 & 0 \\ 3 & -8 & -8 & 1 & 0 \end{bmatrix}\)

A <- matrix(c(1, -2, -1, -1, 0, -5, 11, 7, 4, 1, -22, 49, 33, 16, 0, 3, -8, -8, 1, 0), 4, 5, byrow = TRUE)
part2 <- rref(A)[,5]

For x{0,0,1,0}:

\(\begin{bmatrix} 1 & -2 & -1 & -1 \\ -5 & 11 & 7 & 4 \\ -22 & 49 & 33 & 16 \\ 3 & -8 & -8 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}\)

\(\begin{bmatrix} 1 & -2 & -1 & -1 & 0\\ -5 & 11 & 7 & 4 & 1 \\ -22 & 49 & 33 & 16 & 0 \\ 3 & -8 & -8 & 1 & 0 \end{bmatrix}\)

A <- matrix(c(1, -2, -1, -1, 0, -5, 11, 7, 4, 0, -22, 49, 33, 16, 1, 3, -8, -8, 1, 0), 4, 5, byrow = TRUE)
part3 <- rref(A)[,5]

For x{0,0,0,1}:

\(\begin{bmatrix} 1 & -2 & -1 & -1 \\ -5 & 11 & 7 & 4 \\ -22 & 49 & 33 & 16 \\ 3 & -8 & -8 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\)

\(\begin{bmatrix} 1 & -2 & -1 & -1 & 0 \\ -5 & 11 & 7 & 4 & 0 \\ -22 & 49 & 33 & 16 & 1 \\ 3 & -8 & -8 & 1 & 0 \end{bmatrix}\)

A <- matrix(c(1, -2, -1, -1, 0, -5, 11, 7, 4, 0, -22, 49, 33, 16, 0, 3, -8, -8, 1, 1), 4, 5, byrow = TRUE)
part4 <- rref(A)[,5]

These are the vectors found during this step in a matrix where each column is one vector.

matrix(c(part1, part2, part3, part4), 4, 4, byrow = FALSE)
##      [,1] [,2] [,3] [,4]
## [1,]   20   17   -3    0
## [2,]    7   14   -3   -1
## [3,]    1   -8    2    1
## [4,]    4   -3    1    1

Step 3

Create the formula to solve the problem. Our final expression will be in the form ρB(a + bx + \(cx^2\) + \(dx^3\)).

This formula is equivalent to aρB(1) + bρB(x) + cρB(\(x^2\)) + dρB(\(x^3\)).

a \(\begin{bmatrix} 20 \\ 7 \\ 1 \\ 4 \end{bmatrix}\) + b \(\begin{bmatrix} 17 \\ 14 \\ -8 \\ -3 \end{bmatrix}\) + c \(\begin{bmatrix} -3 \\ -3 \\ 2 \\ 1 \end{bmatrix}\) + d \(\begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix}\)

Using the vectors solved before, we can see the final result.

\(\begin{bmatrix} 20a + 17b - 3c \\ a + 14b - 3c - d \\ a - 8b + 2c + d \\ 4a - 3b + c + d \end{bmatrix}\)

This is our formula.