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Pre-Class Assignment: Matrix Spaces¶
Goals for today's pre-class assignment¶
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%matplotlib inline
import matplotlib.pylab as plt
import numpy as np
import sympy as sym
sym.init_printing(use_unicode=True)
1. Review the Properties of Invertible Matrices¶
Let $A$ be an $n \times n$ matrix. The following statements are equivalent.
- The column vectors of $A$ form a basis for $R^n$
- $|A| \ne 0$
- $A$ is invertible.
- $A$ is row equivalent to $I_n$ (i.e. it's reduced row echelon form is $I_n$)
- The system of equations $Ax = b$ has a unique solution.
- $rank(A) = n$
Consider the following example. We claim that the following set of vectors form a baiss for $R^3$:
$$B = \{(2,1, 4), (-1,6, 0), (2, 4, -3) \}$$Remember for these two vectors to be a basis they need to obay the following two properties:
- They must span $R^3$.
- They must be linearly independent.
Using the above statements we can show this is true in multiple ways.
The column vectors of $A$ form a basis for $R^n$¶
✅ **DO THIS:** Define a numpy matrix A consisting of the vectors $B$ as columns:
#Put your answer to the above question here
from answercheck import checkanswer
checkanswer.matrix(A,'94827a40ec59c7d767afe6841e1723ce');
$|A| \ne 0$¶
✅ **DO THIS:** The first in the above properties tell us that if the vectors in $B$ are truly a basis of $R^3$ then $|A|=0$. Calculate the determinate of $A$ and store the value in det.
#Put your answer to the above question here
#Verify that the determinate is in fact zero
if np.isclose(det,0):
print("Since the Determinate is zero the column vectors do NOT form a Basis")
else:
print("Since the Determinate is non-zero then the column vectors form a Basis.")
$A$ is invertible.¶
✅ **DO THIS:** Since the determinate is non-zero we know that there is an inverse to A. Use python to calculate that inverse and store it in a matrix called A_inv
#put your answer to the above question here
from answercheck import checkanswer
checkanswer.matrix(A_inv,'001aaddd4824f42ad9d2ccde21cf9d24');
$A$ is row equivalent to $I_n$ (i.e. it's reduced row echelon form is $I_n$)¶
✅ **DO THIS:** According to the property above the reduced row echelon form of an invertable matrix is the Identiy matrix. Verify using the python sympy library and store the reduced row echelone matrix in a variable called rref if you really need to check it.
#put your answer to the above question here
from answercheck import checkanswer
checkanswer.matrix(rref,'cde432847c1c4b6d17cd7bfacc457ed1');
The system of equations $Ax = b$ has a unique solution.¶
Let us assume some arbitrary vector $b \in R^n$. According to the above properties it should only have one solution.
✅ **DO THIS:** Find the solution to $Ax=b$ for the vector $b=(-10,200,3)$. Store the solution in a variable called x
from answercheck import checkanswer
checkanswer.vector(x,'161cfd16545b1b5fb13e35d2800f13df');
$rank(A) = n$¶
The final property says that the Rank should equal the dimension of $R^n$. In our example $n=3$. Find a python function to calculate the rank of $A$. Store the value in a variable named rank to check your answer.
#Put your answer to the above quesion here
#Verify that the determinate is in fact zero
if np.isclose(rank,3):
print("Rank is 3")
else:
print("Rank is not 3. Did we do something wrong?")
✅ **QUESTION:** Without doing any calculations (i.e. only using the above properties), how many solutions are there to $Ax=0$? What is(are) the solution(s)?
Put your answer to the above question here.
2. The Basis of a Vector Space¶
Let $U$ be a vector space with basis $B=\{u_1, \ldots, u_n\}$, and let $u$ be a vector in $U$. Because a basis "spans" the vector space, we know that there exists unique scalars $a_1, \dots a_n$ such that:
$$ u = a_1u_1 + \dots + a_nu_n$$The values $a_1, \dots, a_n$ are called the coordinates of $u$ relative to the basis ($B$) and is typically written as a column vector:
$$ u_B = \left[ \begin{matrix} a_1 \\ \vdots \\ a_n \end{matrix} \right] $$We can create a transition matrix $P$ using the inverse of the matrix with the basis vectors being columns.
$$P = [ u_1 \ldots u_n ]^{-1}$$Now we will show that matrix $P$ will transition vector $u$ in the standard coordinate system to the coordinates relative to the basis $B$:
$$ u_B = Pu$$EXAMPLE: Consider the vector $u = \left[ \begin{matrix} 5 \\ 3 \end{matrix} \right]$ and the basis vectors $B = \{(1,2), (3,-1)\}$. The following code calculate the $P$ transition matrix from $B$ and then use $P$ to calculate the values of $u_B$ ($a_1$ and $a_2$):
u = np.matrix([[5],[3]])
sym.Matrix(u)
B = np.matrix([[1,2],[3,-1]]).T
sym.Matrix(B)
P = np.linalg.inv(B)
ub = P*u
sym.Matrix(ub)
Here we would like to view this from $R^n$. Let $$B=[u_1 \dots u_n],$$ then the values of $u_B$ can be found by solving the linear system $$u = B u_B.$$ Because the columns of $B$ is a basis, therefore, the matrix $B$ is a $n\times n$ square matrix and it has an inverse. Therefore, we can solve the linear system and obtain $$u_B = B^{-1} u = Pu.$$
Let's try to visualize this with a plot:
ax = plt.axes();
#Blue arrow representing first Basis Vector
ax.arrow(0, 0, B[0,0],B[1,0], head_width=.2, head_length=.2, fc='blue', ec='blue');
#Green arrow representing Second Basis Vector
plt.plot([0,B[0,1]],[0,B[1,1]],color='green'); #Need this line to make the figure work. Not sure why.
ax.arrow(0, 0, B[0,1],B[1,1], head_width=.2, head_length=.2, fc='green', ec='green');
#Original point u as a red dot
ax.scatter(u[0,0],u[1,0], color='red');
plt.show()
#plt.axis('equal');
Notice that the blue arrow represents the first basis vector and the green arrow is the second basis vector in $B$. The solution to $u_B$ shows 2 units along the blue vector and 1 units along the green vector, which puts us at the point (5,3).
This is also called a change in coordinate systems.
✅ **QUESTION**: What is the coordinate vector of $u$ relative to the given basis $B$ in $R^3$?
$$u = (9,-3,21)$$$$B = \{(2,0,-1), (0,1,3), (1,1,1)\}$$Store this coordinate in a variable ub for checking:
#Put your answer here
from answercheck import checkanswer
checkanswer.vector(ub,'f72f62c739096030e0074c4e1dfca159');
Let's look more closely into the matrix $P$, what is the meaning of the columns of the matrix $P$?
We know that $P$ is the inverse of $B$, therefore, we have $$BP=I.$$ Then we can look at the first column of the $P$, say $p_{1}$, we have that $Bp_1$ is the column vector $(1,0,0)^\top$, which is exactly the first component from the standard basis. This is true for other columns.
It means that if we want to change an old basis $B$ to a new basis $B'$, we need to find out all the coordinates in the new basis for the old basis, and the transition matrix is by putting all the coordinates as columns.
Here is the matrix $B$ again:
B = np.matrix([[2,0,-1],[0,1,3],[1,1,1]]).T
sym.Matrix(B)
The first column of P should be the solution to $Bx=\left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right]$. We can use the numpy.linalg.solve function to find this solution:
# The first column of P should be
u1 = np.matrix([1,0,0]).T
p1 = np.linalg.solve(B,u1)
p1
We can find a similar answer for columns $p_2$ and $p_3$:
# The second column of P should be
u2 = np.matrix([0,1,0]).T
p2 = np.linalg.solve(B,u2)
p2
# The third column of P should be
u3 = np.matrix([0,0,1]).T
p3 = np.linalg.solve(B,u3)
p3
# concatenate three column together into a 3x3 matrix
P = np.concatenate((p1, p2, p3), axis=1)
sym.Matrix(P)
# Find the new coordinate in the new basis
u = np.matrix([9,-3,21]).T
UB = P*u
print(UB)
This should be basically the same answer as you got above.
3. Change of Basis¶
Now consider the following two bases in $R^2$:
$$B_1 = \{(1,2), (3,-1)\}$$$$B_2 = \{(3,1), (5,2)\}$$The transformation from the "standard basis" to $B_1$ and $B_2$ can be defined as the column vectors $P_1$ and $P_2$ as follows:
B1 = np.matrix([[1,2],[3,-1]]).T
P1 = np.linalg.inv(B1)
sym.Matrix(P1)
B2 = np.matrix([[3,1],[5,2]]).T
P2 = np.linalg.inv(B2)
sym.Matrix(P2)
✅ **DO THIS**: Find the transition matrix $T$ that will take points in the $B_1$ coordinate representation and put them into $B_2$ coordinates. NOTE this is analogous to the robot kinematics problem. We want to represent points in a different coordinate system.
# Put your answer to the above question here.
from answercheck import checkanswer
checkanswer.matrix(T,'dcc03ddff982e29eea6dd52ec9088986')
✅ **QUESTION 6**: Given $u_{B_1} = \left[ \begin{matrix} 2 \\ 1 \end{matrix} \right]$ (a point named $u$ in the $B_1$ coordinate system) and your calculated transition matrix $T$, what is the same point expressed in the $B_2$ basis (i.e. what is $u_{B2}$)? Store your answer in a variable named ub2 for checking.
ub1 = np.matrix([[2],[1]])
sym.Matrix(ub1)
##Put your code here
from answercheck import checkanswer
checkanswer.vector(ub2,'9a5fe29254c07cf59ebdffcaba679917')
There are three bases $B_1$, $B_2$, and $B_3$. We have the transition matrix $P_{12}$ from $B_1$ to $B_2$ and the transition matrix $P_{23}$ from $B_2$ to $B_3$. In $R^n$, we can compute the transition matrix as $$P_{12}=B_2^{-1}B_1,\quad P_{23}=B_3^{-1}B_2$$
Then we can find all other transition matrices. $$P_{13} = B_3^{-1}B_1=B_3^{-1}B_2*B_2^{-1}B_1= P_{23}P_{12}$$ $$P_{21} = B_1^{-1}B_2 = (B_2^{-1}B_1)^{-1}=P_{12}^{-1}$$ $$P_{32} = B_2^{-1}B_3 = (B_3^{-1}B_2)^{-1}=P_{23}^{-1}$$ $$P_{31} = B_1^{-1}B_3 = (B_3^{-1}B_1)^{-1}=P_{13}^{-1}=(P_{23}P_{12})^{-1}=P_{12}^{-1}P_{23}^{-1}$$
The result is true for general vector spaces and can be extended to many bases.
4. Assignment wrap-up¶
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Written by Dirk Colbry, Michigan State University
This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.