4228-Section 1.8.sagews

Section 1.8

#1a. Find the image of u=(1,-3) under the transformation A.

u = vector(QQ,[1,-3])
A = matrix(QQ,[[2,0],[0,2]])
A*u.column()

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{r} 2 \\ -6 \end{array}$$\right)

# Here is one way to plot a vector
u = arrow((0,0),(1,-3), color='red')
Au = arrow((0,0),(2,-6), color='blue')
title = text("Vector u (red), and its image Au (blue).", (1.5,-1))
plot(u+Au+title)

#5.

\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left($$\begin{array}{rrr} 1 & -5 & -7 \\ -3 & 7 & 5 \end{array}$$\right), \left(-2,\,-2\right)\right)

Augment b to get [A b]

Ab = A.augment(b)
Ab

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & -5 & -7 & -2 \\ -3 & 7 & 5 & -2 \end{array}$$\right)

Ab.rref()

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & 0 & 3 & 3 \\ 0 & 1 & 2 & 1 \end{array}$$\right)

Thus, x_1 = 3-3x_3, and x_2 = 1-2x_3. The solution is not unique since x_3 is free! The general solution is the following:

\newcommand{\Bold}[1]{\mathbf{#1}}\left(-3 \, x_{3} + 3,\,-x_{3} + 1,\,x_{3}\right)

An easy-to-find particular solution results from setting x_3 = 0:

\newcommand{\Bold}[1]{\mathbf{#1}}\left(3,\,1,\,0\right)

# 9 (Alternate solution from the future (see Section 4.2). The kernel is the set {x} such that Ax=0. For now, this problem should be solved using row reduction algorithms. This is designed to give you a sneak peak at the general notions in this problem.)

A = matrix(QQ,[[1,-3,5,-5],[0,1,-3,5],[2,-4,4,-4]])
A

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & -3 & 5 & -5 \\ 0 & 1 & -3 & 5 \\ 2 & -4 & 4 & -4 \end{array}$$\right)

A.right_kernel()

\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{RowSpan}_{\Bold{Q}}\left($$\begin{array}{rrrr} 1 & \frac{3}{4} & \frac{1}{4} & 0 \end{array}$$\right)

The span of the vector (1,3/4,1/4,0) forms the kernel, or "null space", of the transformation A.