12 Pre-Class Assignment: Matrix Spaces¶

Readings for this topic (Recommended in bold)¶

Goals for today’s pre-class assignment¶

Properties of Invertible Matrices
The Basis of a Vector Space
Change of Basis
Assignment wrap-up

%matplotlib inline
import matplotlib.pylab as plt
import numpy as np
import sympy as sym
sym.init_printing(use_unicode=True)

---------------------------------------------------------------------------
ModuleNotFoundError                       Traceback (most recent call last)
<ipython-input-1-e3122a161773> in <module>
----> 1 get_ipython().run_line_magic('matplotlib', 'inline')
      2 import matplotlib.pylab as plt
      3 import numpy as np
      4 import sympy as sym
      5 sym.init_printing(use_unicode=True)

~/REPOS/MTH314_Textbook/MakeTextbook/envs/lib/python3.9/site-packages/IPython/core/interactiveshell.py in run_line_magic(self, magic_name, line, _stack_depth)
   2342                 kwargs['local_ns'] = self.get_local_scope(stack_depth)
   2343             with self.builtin_trap:
-> 2344                 result = fn(*args, **kwargs)
   2345             return result
   2346 

~/REPOS/MTH314_Textbook/MakeTextbook/envs/lib/python3.9/site-packages/decorator.py in fun(*args, **kw)
    230             if not kwsyntax:
    231                 args, kw = fix(args, kw, sig)
--> 232             return caller(func, *(extras + args), **kw)
    233     fun.__name__ = func.__name__
    234     fun.__doc__ = func.__doc__

~/REPOS/MTH314_Textbook/MakeTextbook/envs/lib/python3.9/site-packages/IPython/core/magic.py in <lambda>(f, *a, **k)
    185     # but it's overkill for just that one bit of state.
    186     def magic_deco(arg):
--> 187         call = lambda f, *a, **k: f(*a, **k)
    188 
    189         if callable(arg):

~/REPOS/MTH314_Textbook/MakeTextbook/envs/lib/python3.9/site-packages/IPython/core/magics/pylab.py in matplotlib(self, line)
     97             print("Available matplotlib backends: %s" % backends_list)
     98         else:
---> 99             gui, backend = self.shell.enable_matplotlib(args.gui.lower() if isinstance(args.gui, str) else args.gui)
    100             self._show_matplotlib_backend(args.gui, backend)
    101 

~/REPOS/MTH314_Textbook/MakeTextbook/envs/lib/python3.9/site-packages/IPython/core/interactiveshell.py in enable_matplotlib(self, gui)
   3511         """
   3512         from IPython.core import pylabtools as pt
-> 3513         gui, backend = pt.find_gui_and_backend(gui, self.pylab_gui_select)
   3514 
   3515         if gui != 'inline':

~/REPOS/MTH314_Textbook/MakeTextbook/envs/lib/python3.9/site-packages/IPython/core/pylabtools.py in find_gui_and_backend(gui, gui_select)
    278     """
    279 
--> 280     import matplotlib
    281 
    282     if gui and gui != 'auto':

ModuleNotFoundError: No module named 'matplotlib'

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 basis 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 determinant 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 determinant 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 Identity 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 (assignment-specific): 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 scalars $a_1, \dots, a_n$ such that:

\[ u = a_1u_1 + \dots + a_nu_n\]

Since a basis is a linearly independent set of vectors we know the scalars $a_1, \dots, a_n$ are unique.

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:

\[\begin{split} u_B = \left[ \begin{matrix} a_1 \\ \vdots \\ a_n \end{matrix} \right] \end{split}\]

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 calculates the $P$ transition matrix from $B$ and then uses $P$ 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 $$u = B u_B.$$ The columns of $B$ are 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: 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¶

✅ Assignment-Specific 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)?