Math211.1.8

108 days ago by Jesse_Berwald

%hide %html <h3>Section 1.8</h3> 
       

Section 1.8

Section 1.8
%hide %html #1a. Find the image of u=(1,-3) under the transformation A. 
       
#1a. Find the image of u=(1,-3) under the transformation A.
#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)
\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) 
       
%hide %html #5. 
       
#5.
#5.
%hide A = matrix(QQ,[[1,-5,-7],[-3,7,5]]) b = vector(QQ,[-2,-2]) A,b 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left(\begin{array}{rrr}
1 & -5 & -7 \\
-3 & 7 & 5
\end{array}\right), \left(-2,\,-2\right)\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left(\begin{array}{rrr}
1 & -5 & -7 \\
-3 & 7 & 5
\end{array}\right), \left(-2,\,-2\right)\right)
%hide %html Augment b to get [A b] 
       
Augment b to get [A b]
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)
\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)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr}
1 & 0 & 3 & 3 \\
0 & 1 & 2 & 1
\end{array}\right)
%hide %html 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: 
       
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:
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:
%hide x_3 = var('x_3') G = vector(SR,[3-3*x_3,1-x_3,x_3]); G 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left(-3 \, x_{3} + 3,\,-x_{3} + 1,\,x_{3}\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(-3 \, x_{3} + 3,\,-x_{3} + 1,\,x_{3}\right)
%hide %html An easy-to-find particular solution results from setting x_3 = 0: 
       
An easy-to-find particular solution results from setting x_3 = 0:
An easy-to-find particular solution results from setting x_3 = 0:
%hide vector(SR,[3,1,0]) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left(3,\,1,\,0\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(3,\,1,\,0\right)
%hide %html # 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.) 
       
# 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.)
# 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)
\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)
\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{RowSpan}_{\Bold{Q}}\left(\begin{array}{rrrr}
1 & \frac{3}{4} & \frac{1}{4} & 0
\end{array}\right)
%hide %html The span of the vector (1,3/4,1/4,0) forms the kernel, or "null space", of the transformation A. 
       
The span of the vector (1,3/4,1/4,0) forms the kernel, or "null space", of the transformation A.
The span of the vector (1,3/4,1/4,0) forms the kernel, or "null space", of the transformation A.