Tutorial 10 (AY24/25 Sem 1)¶

In [1]:

# Required imports
import sympy as sym
from ma1522 import Matrix

# Required imports import sympy as sym from ma1522 import Matrix

Question 1¶

A population of ants is put into a maze with 3 compartments labeled $a$, $b$, and $c$. If the ant is in compartment $a$, an hour later, there is a 20% chance it will go to compartment $b$, and a 40% chance it will go to compartment $c$. If it is in compartment $b$, an hour later, there is a 10% chance it will go to compartment $a$, and a 30% chance it will go to compartment $c$. If it is in compartment $c$, an hour later, there is a 50% chance it will go to compartment $a$, and a 20% chance it will go to compartment $b$. Suppose 100 ants have been placed in compartment $a$.

(a)¶

Find the transition probability matrix $\mathbf{A}$. Show that it is a stochastic matrix.

In [2]:

A = Matrix.from_str("0.4 0.1 0.5; 0.2 0.6 0.2; 0.4 0.3 0.3")
A

A = Matrix.from_str("0.4 0.1 0.5; 0.2 0.6 0.2; 0.4 0.3 0.3") A

Out[2]:

$\displaystyle \left[\begin{array}{ccc}0.4 & 0.1 & 0.5\\0.2 & 0.6 & 0.2\\0.4 & 0.3 & 0.3\end{array}\right]$

In [3]:

# Always work with fractions to avoid floating point errors
A.simplify(rational=True) 
A

# Always work with fractions to avoid floating point errors A.simplify(rational=True) A

Out[3]:

$\displaystyle \left[\begin{array}{ccc}\frac{2}{5} & \frac{1}{10} & \frac{1}{2}\\\frac{1}{5} & \frac{3}{5} & \frac{1}{5}\\\frac{2}{5} & \frac{3}{10} & \frac{3}{10}\end{array}\right]$

In [4]:

A.is_stochastic(verbosity=1)

A.is_stochastic(verbosity=1)

Check if matrix is square: True
Check if all entries are non-negative: True
Check if each column sums to 1: True

Out[4]:

True

(b)¶

By diagonalizing $\mathbf{A}$, find the number of ants in each compartment after 3 hours.

In [5]:

PDP = A.diagonalize(reals_only=True, verbosity=1)
PDP

PDP = A.diagonalize(reals_only=True, verbosity=1) PDP

Characteristic Polynomial

$\displaystyle \left(x - 1\right) \left(x - \frac{2}{5}\right) \left(x + \frac{1}{10}\right)$

$\displaystyle \text{Before RREF: }1\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}\frac{3}{5} & - \frac{1}{10} & - \frac{1}{2}\\- \frac{1}{5} & \frac{2}{5} & - \frac{1}{5}\\- \frac{2}{5} & - \frac{3}{10} & \frac{7}{10}\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & -1\\0 & 1 & -1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}1\\1\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }\frac{2}{5}\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}0 & - \frac{1}{10} & - \frac{1}{2}\\- \frac{1}{5} & - \frac{1}{5} & - \frac{1}{5}\\- \frac{2}{5} & - \frac{3}{10} & \frac{1}{10}\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & -4\\0 & 1 & 5\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}4\\-5\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }- \frac{1}{10}\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}- \frac{1}{2} & - \frac{1}{10} & - \frac{1}{2}\\- \frac{1}{5} & - \frac{7}{10} & - \frac{1}{5}\\- \frac{2}{5} & - \frac{3}{10} & - \frac{2}{5}\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 0\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\0\\1\end{matrix}\right]\right]$

Out[5]:

$\left[\begin{array}{ccc}-1 & 4 & 1\\0 & -5 & 1\\1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}- \frac{1}{10} & 0 & 0\\0 & \frac{2}{5} & 0\\0 & 0 & 1\end{array}\right]\left[\begin{matrix}- \frac{2}{5} & - \frac{1}{5} & \frac{3}{5}\\\frac{1}{15} & - \frac{2}{15} & \frac{1}{15}\\\frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{matrix}\right]$

In [6]:

x_0 = Matrix.from_str("100; 0; 0")
x_0

x_0 = Matrix.from_str("100; 0; 0") x_0

Out[6]:

$\displaystyle \left[\begin{array}{c}100\\0\\0\end{array}\right]$

In [7]:

x_3 = PDP.P @ PDP.D**3 @ PDP.P.inv() @ x_0
x_3

x_3 = PDP.P @ PDP.D**3 @ PDP.P.inv() @ x_0 x_3

Out[7]:

$\displaystyle \left[\begin{array}{c}35\\\frac{156}{5}\\\frac{169}{5}\end{array}\right]$

In [8]:

x_3.evalf()

x_3.evalf()

Out[8]:

$\displaystyle \left[\begin{matrix}35.0\\31.2\\33.8\end{matrix}\right]$

(d)¶

In the long run (assuming no ants died), where will the majority of the ants be?

In [9]:

# Brute force calculation
(A**100 @ x_0).evalf()

# Brute force calculation (A**100 @ x_0).evalf()

Out[9]:

$\displaystyle \left[\begin{matrix}33.3333333333333\\33.3333333333333\\33.3333333333333\end{matrix}\right]$

In [10]:

D_100 = (PDP.D**1000).evalf()
D_100

D_100 = (PDP.D**1000).evalf() D_100

Out[10]:

$\displaystyle \left[\begin{matrix}1.0 \cdot 10^{-1000} & 0 & 0\\0 & 1.14813069527425 \cdot 10^{-398} & 0\\0 & 0 & 1.0\end{matrix}\right]$

In [11]:

# Therefore,
D_inf = Matrix.from_str("0 0 0; 0 0 0; 0 0 1")
D_inf

# Therefore, D_inf = Matrix.from_str("0 0 0; 0 0 0; 0 0 1") D_inf

Out[11]:

$\displaystyle \left[\begin{array}{ccc}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 1\end{array}\right]$

In [12]:

PDP.P @ D_inf @ PDP.P.inv() @ x_0

PDP.P @ D_inf @ PDP.P.inv() @ x_0

Out[12]:

$\displaystyle \left[\begin{array}{c}\frac{100}{3}\\\frac{100}{3}\\\frac{100}{3}\end{array}\right]$

(e)¶

Suppose initially the numbers of ants in compartments a, b and c are $\alpha$, $\beta$, and $\gamma$ respectively. What is the population distribution in the long run (assuming no ants died)?

In [13]:

# simple random vector
Matrix.create_unk_matrix(3, 1)

# simple random vector Matrix.create_unk_matrix(3, 1)

Out[13]:

$\displaystyle \left[\begin{array}{c}x\\y\\z\end{array}\right]$

In [14]:

# To use custom symbols, the most straightforward way is
# to use sympy's symbols function
alpha, beta, gamma = sym.symbols("alpha beta gamma")
x = Matrix([alpha, beta, gamma])
x

# To use custom symbols, the most straightforward way is # to use sympy's symbols function alpha, beta, gamma = sym.symbols("alpha beta gamma") x = Matrix([alpha, beta, gamma]) x

Out[14]:

$\displaystyle \left[\begin{array}{c}\alpha\\\beta\\\gamma\end{array}\right]$

In [15]:

PDP.P @ D_inf @ PDP.P.inv() @ x

PDP.P @ D_inf @ PDP.P.inv() @ x

Out[15]:

$\displaystyle \left[\begin{array}{c}\frac{\alpha}{3} + \frac{\beta}{3} + \frac{\gamma}{3}\\\frac{\alpha}{3} + \frac{\beta}{3} + \frac{\gamma}{3}\\\frac{\alpha}{3} + \frac{\beta}{3} + \frac{\gamma}{3}\end{array}\right]$

Question 2¶

By diagonalizing $\mathbf{A} = \begin{pmatrix} 1 & 0 & 3 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix}$, find a matrix $\mathbf{B}$ such that $\mathbf{B}^2 = \mathbf{A}$.

In [16]:

A = Matrix.from_str("1 0 3; 0 4 0; 0 0 4")
A

A = Matrix.from_str("1 0 3; 0 4 0; 0 0 4") A

Out[16]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 3\\0 & 4 & 0\\0 & 0 & 4\end{array}\right]$

In [17]:

PDP = A.diagonalize(reals_only=True, verbosity=1)
PDP

PDP = A.diagonalize(reals_only=True, verbosity=1) PDP

Characteristic Polynomial

$\displaystyle \left(x - 4\right)^{2} \left(x - 1\right)$

$\displaystyle \text{Before RREF: }1\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}0 & 0 & -3\\0 & -3 & 0\\0 & 0 & -3\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (1, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}1\\0\\0\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }4\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}3 & 0 & -3\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & -1\\0 & 0 & 0\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0,)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}0\\1\\0\end{matrix}\right], \ \left[\begin{matrix}1\\0\\1\end{matrix}\right]\right]$

Out[17]:

$\left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0\\0 & 4 & 0\\0 & 0 & 4\end{array}\right]\left[\begin{matrix}1 & 0 & -1\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right]$

In [18]:

# Therefore, A = B**2 --> B = PDP.P @ PDP.D**0.5 @ PDP.P.inv()
# Note: sym.sqrt only produces the principal square root
B = PDP.P @ sym.sqrt(PDP.D) @ PDP.P.inv() 
B

# Therefore, A = B**2 --> B = PDP.P @ PDP.D**0.5 @ PDP.P.inv() # Note: sym.sqrt only produces the principal square root B = PDP.P @ sym.sqrt(PDP.D) @ PDP.P.inv() B

Out[18]:

$\displaystyle \left[\begin{matrix}1 & 0 & 1\\0 & 2 & 0\\0 & 0 & 2\end{matrix}\right]$

Problem 3¶

For each of the following symmetric matrices $\mathbf{A}$, find an orthogonal matrix $\mathbf{P}$ that orthogonally diagonalizes $\mathbf{A}$.

(a)¶

$\mathbf{A} = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$

In [19]:

A = Matrix.from_str("3 1; 1 3")
A

A = Matrix.from_str("3 1; 1 3") A

Out[19]:

$\displaystyle \left[\begin{array}{cc}3 & 1\\1 & 3\end{array}\right]$

In [20]:

A.orthogonally_diagonalize(verbosity=1)

A.orthogonally_diagonalize(verbosity=1)

Check if matrix is symmetric: True
Characteristic Polynomial

$\displaystyle \left(x - 4\right) \left(x - 2\right)$

$\displaystyle \text{Before RREF: }4\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cc}1 & -1\\-1 & 1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cc}1 & -1\\0 & 0\end{array}\right], \quad\text{pivots} = (0,)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}1\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }2\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cc}-1 & -1\\-1 & -1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cc}1 & 1\\0 & 0\end{array}\right], \quad\text{pivots} = (0,)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\1\end{matrix}\right]\right]$

Out[20]:

$\left[\begin{array}{cc}- \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{array}\right]\left[\begin{array}{cc}2 & 0\\0 & 4\end{array}\right]\left[\begin{matrix}- \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix}\right]$

(b)¶

$\mathbf{A} = \begin{pmatrix} 2 & 2 & -2 \\ 2 & -1 & 4 \\ -2 & 4 & -1 \end{pmatrix}$

In [21]:

A = Matrix.from_str("2 2 -2; 2 -1 4; -2 4 -1")
A

A = Matrix.from_str("2 2 -2; 2 -1 4; -2 4 -1") A

Out[21]:

$\displaystyle \left[\begin{array}{ccc}2 & 2 & -2\\2 & -1 & 4\\-2 & 4 & -1\end{array}\right]$

In [22]:

PDPT = A.orthogonally_diagonalize(verbosity=1)
PDPT

PDPT = A.orthogonally_diagonalize(verbosity=1) PDPT

Check if matrix is symmetric: True
Characteristic Polynomial

$\displaystyle \left(x - 3\right)^{2} \left(x + 6\right)$

$\displaystyle \text{Before RREF: }-6\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}-8 & -2 & 2\\-2 & -5 & -4\\2 & -4 & -5\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & - \frac{1}{2}\\0 & 1 & 1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}\frac{1}{2}\\-1\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }3\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}1 & -2 & 2\\-2 & 4 & -4\\2 & -4 & 4\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & -2 & 2\\0 & 0 & 0\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0,)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}2\\1\\0\end{matrix}\right], \ \left[\begin{matrix}-2\\0\\1\end{matrix}\right]\right]$

Eigenvalue:  3
[Gram Schmidt Process]

$\displaystyle v_1 = \left[\begin{array}{c}2\\1\\0\end{array}\right]$

$\displaystyle v_2 = \left[\begin{array}{c}-2\\0\\1\end{array}\right]- \left(\frac{-4}{5}\right) \left[\begin{array}{c}2\\1\\0\end{array}\right] = \left(\frac{1}{5}\right) \left[\begin{matrix}-2\\4\\5\end{matrix}\right]$

Out[22]:

$\left[\begin{array}{ccc}\frac{1}{3} & \frac{2 \sqrt{5}}{5} & - \frac{2 \sqrt{5}}{15}\\- \frac{2}{3} & \frac{\sqrt{5}}{5} & \frac{4 \sqrt{5}}{15}\\\frac{2}{3} & 0 & \frac{\sqrt{5}}{3}\end{array}\right]\left[\begin{array}{ccc}-6 & 0 & 0\\0 & 3 & 0\\0 & 0 & 3\end{array}\right]\left[\begin{matrix}\frac{1}{3} & - \frac{2}{3} & \frac{2}{3}\\\frac{2 \sqrt{5}}{5} & \frac{\sqrt{5}}{5} & 0\\- \frac{2 \sqrt{5}}{15} & \frac{4 \sqrt{5}}{15} & \frac{\sqrt{5}}{3}\end{matrix}\right]$

Question 4¶

Let $\mathbf{A} = \begin{pmatrix} 1 & -2 & 0 & 0 \\ -2 & 1 & 0 & 0 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & -2 & 1 \end{pmatrix}$.

(a)¶

Find an orthogonal matrix $\mathbf{P}$ that orthogonally diagonalizes $\mathbf{A}$, and compute $\mathbf{P}^T \mathbf{A} \mathbf{P}$.

In [23]:

A = Matrix.from_str("1 -2 0 0; -2 1 0 0; 0 0 1 -2; 0 0 -2 1")
A

A = Matrix.from_str("1 -2 0 0; -2 1 0 0; 0 0 1 -2; 0 0 -2 1") A

Out[23]:

$\displaystyle \left[\begin{array}{cccc}1 & -2 & 0 & 0\\-2 & 1 & 0 & 0\\0 & 0 & 1 & -2\\0 & 0 & -2 & 1\end{array}\right]$

In [24]:

PDPT = A.orthogonally_diagonalize(verbosity=1)
PDPT

PDPT = A.orthogonally_diagonalize(verbosity=1) PDPT

Check if matrix is symmetric: True
Characteristic Polynomial

$\displaystyle \left(x - 3\right)^{2} \left(x + 1\right)^{2}$

$\displaystyle \text{Before RREF: }3\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}2 & 2 & 0 & 0\\2 & 2 & 0 & 0\\0 & 0 & 2 & 2\\0 & 0 & 2 & 2\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 1 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\1\\0\\0\end{matrix}\right], \ \left[\begin{matrix}0\\0\\-1\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }-1\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}-2 & 2 & 0 & 0\\2 & -2 & 0 & 0\\0 & 0 & -2 & 2\\0 & 0 & 2 & -2\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & -1 & 0 & 0\\0 & 0 & 1 & -1\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}1\\1\\0\\0\end{matrix}\right], \ \left[\begin{matrix}0\\0\\1\\1\end{matrix}\right]\right]$

Eigenvalue:  -1
[Gram Schmidt Process]

$\displaystyle v_1 = \left[\begin{array}{c}1\\1\\0\\0\end{array}\right]$

$\displaystyle v_2 = \left[\begin{array}{c}0\\0\\1\\1\end{array}\right]- \left(\frac{0}{2}\right) \left[\begin{array}{c}1\\1\\0\\0\end{array}\right] = 1 \left[\begin{matrix}0\\0\\1\\1\end{matrix}\right]$

Eigenvalue:  3
[Gram Schmidt Process]

$\displaystyle v_1 = \left[\begin{array}{c}-1\\1\\0\\0\end{array}\right]$

$\displaystyle v_2 = \left[\begin{array}{c}0\\0\\-1\\1\end{array}\right]- \left(\frac{0}{2}\right) \left[\begin{array}{c}-1\\1\\0\\0\end{array}\right] = 1 \left[\begin{matrix}0\\0\\-1\\1\end{matrix}\right]$

Out[24]:

$\left[\begin{array}{cccc}\frac{\sqrt{2}}{2} & 0 & - \frac{\sqrt{2}}{2} & 0\\\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} & 0\\0 & \frac{\sqrt{2}}{2} & 0 & - \frac{\sqrt{2}}{2}\\0 & \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{array}\right]\left[\begin{array}{cccc}-1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & 3 & 0\\0 & 0 & 0 & 3\end{array}\right]\left[\begin{matrix}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 & 0\\0 & 0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\\- \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 & 0\\0 & 0 & - \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix}\right]$

Question 5¶

Find the SVD of the following matrices $\mathbf{A}$.

(a)¶

$\mathbf{A} = \begin{pmatrix} 3 & 2 \\ 2 & 3 \\ 2 & -2 \end{pmatrix}$

In [25]:

A = Matrix.from_str("3 2; 2 3; 2 -2")
A

A = Matrix.from_str("3 2; 2 3; 2 -2") A

Out[25]:

$\displaystyle \left[\begin{array}{cc}3 & 2\\2 & 3\\2 & -2\end{array}\right]$

In [26]:

svd = A.singular_value_decomposition(verbosity=1)
svd

svd = A.singular_value_decomposition(verbosity=1) svd

A^T A

$\displaystyle \left[\begin{array}{cc}17 & 8\\8 & 17\end{array}\right]$

Check if matrix is symmetric: True
Characteristic Polynomial

$\displaystyle \left(x - 25\right) \left(x - 9\right)$

$\displaystyle \text{Before RREF: }25\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cc}8 & -8\\-8 & 8\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cc}1 & -1\\0 & 0\end{array}\right], \quad\text{pivots} = (0,)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}1\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }9\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cc}-8 & -8\\-8 & -8\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cc}1 & 1\\0 & 0\end{array}\right], \quad\text{pivots} = (0,)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\1\end{matrix}\right]\right]$

$\displaystyle u_1 = (1/5)A\left[\begin{matrix}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] = \left[\begin{array}{c}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\\0\end{array}\right]$

$\displaystyle u_2 = (1/3)A\left[\begin{matrix}- \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] = \left[\begin{array}{c}- \frac{\sqrt{2}}{6}\\\frac{\sqrt{2}}{6}\\- \frac{2 \sqrt{2}}{3}\end{array}\right]$

Extending U with its orthogonal complement.
Before RREF: [self]

$\displaystyle \left[\begin{matrix}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\\- \frac{\sqrt{2}}{6} & \frac{\sqrt{2}}{6} & - \frac{2 \sqrt{2}}{3}\end{matrix}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & 2\\0 & 1 & -2\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

$\displaystyle v_1 = \left[\begin{array}{c}-2\\2\\1\end{array}\right]$

Out[26]:

$\left[\begin{array}{ccc}\frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{6} & - \frac{2}{3}\\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{6} & \frac{2}{3}\\0 & - \frac{2 \sqrt{2}}{3} & \frac{1}{3}\end{array}\right]\left[\begin{array}{cc}5 & 0\\0 & 3\\0 & 0\end{array}\right]\left[\begin{array}{cc}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\\- \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{array}\right]$

(b)¶

$\mathbf{A} = \begin{pmatrix} 3 & 2 & 2 \\ 2 & 3 & -2 \end{pmatrix}$

In [27]:

# Import the SVD data class to represent the result of SVD
from ma1522 import SVD 

# Import the SVD data class to represent the result of SVD from ma1522 import SVD

In [28]:

# A = U @ S @ V.T
# A.T = V @ S.T @ U.T
svdT = SVD(U=svd.V, S = svd.S.T, V=svd.U)
svdT

# A = U @ S @ V.T # A.T = V @ S.T @ U.T svdT = SVD(U=svd.V, S = svd.S.T, V=svd.U) svdT

Out[28]:

$\left[\begin{array}{cc}\frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{array}\right]\left[\begin{array}{ccc}5 & 0 & 0\\0 & 3 & 0\end{array}\right]\left[\begin{array}{ccc}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\\- \frac{\sqrt{2}}{6} & \frac{\sqrt{2}}{6} & - \frac{2 \sqrt{2}}{3}\\- \frac{2}{3} & \frac{2}{3} & \frac{1}{3}\end{array}\right]$

In [29]:

svdT.eval()

svdT.eval()

Out[29]:

$\displaystyle \left[\begin{matrix}3 & 2 & 2\\2 & 3 & -2\end{matrix}\right]$

(c)¶

$\mathbf{A} = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}$

In [30]:

A = Matrix.from_str("1 0 1; 0 1 1; 1 1 2")
A

A = Matrix.from_str("1 0 1; 0 1 1; 1 1 2") A

Out[30]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 1\\1 & 1 & 2\end{array}\right]$

In [31]:

A.singular_value_decomposition(verbosity=1)

A.singular_value_decomposition(verbosity=1)

A^T A

$\displaystyle \left[\begin{array}{ccc}2 & 1 & 3\\1 & 2 & 3\\3 & 3 & 6\end{array}\right]$

Check if matrix is symmetric: True
Characteristic Polynomial

$\displaystyle x \left(x - 9\right) \left(x - 1\right)$

$\displaystyle \text{Before RREF: }9\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}7 & -1 & -3\\-1 & 7 & -3\\-3 & -3 & 3\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & - \frac{1}{2}\\0 & 1 & - \frac{1}{2}\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}\frac{1}{2}\\\frac{1}{2}\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }1\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}-1 & -1 & -3\\-1 & -1 & -3\\-3 & -3 & -5\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\1\\0\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }0\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}-2 & -1 & -3\\-1 & -2 & -3\\-3 & -3 & -6\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\-1\\1\end{matrix}\right]\right]$

$\displaystyle u_1 = (1/3)A\left[\begin{matrix}\frac{\sqrt{6}}{6}\\\frac{\sqrt{6}}{6}\\\frac{\sqrt{6}}{3}\end{matrix}\right] = \left[\begin{array}{c}\frac{\sqrt{6}}{6}\\\frac{\sqrt{6}}{6}\\\frac{\sqrt{6}}{3}\end{array}\right]$

$\displaystyle u_2 = (1/1)A\left[\begin{matrix}- \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\\0\end{matrix}\right] = \left[\begin{array}{c}- \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\\0\end{array}\right]$

Extending U with its orthogonal complement.
Before RREF: [self]

$\displaystyle \left[\begin{matrix}\frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{3}\\- \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\end{matrix}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 1\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

$\displaystyle v_1 = \left[\begin{array}{c}-1\\-1\\1\end{array}\right]$

Out[31]:

$\left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & - \frac{\sqrt{2}}{2} & - \frac{\sqrt{3}}{3}\\\frac{\sqrt{6}}{6} & \frac{\sqrt{2}}{2} & - \frac{\sqrt{3}}{3}\\\frac{\sqrt{6}}{3} & 0 & \frac{\sqrt{3}}{3}\end{array}\right]\left[\begin{array}{ccc}3 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{3}\\- \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\\- \frac{\sqrt{3}}{3} & - \frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3}\end{array}\right]$

In [32]:

# OR symmetric -> diagonalize == orthogonally diagonalize == svd
pdp = A.diagonalize(verbosity=1, sort=True)  # This will also work
pdp # Take note that the singular values might not be in the correct order

# OR symmetric -> diagonalize == orthogonally diagonalize == svd pdp = A.diagonalize(verbosity=1, sort=True) # This will also work pdp # Take note that the singular values might not be in the correct order

Characteristic Polynomial

$\displaystyle x \left(x - 3\right) \left(x - 1\right)$

$\displaystyle \text{Before RREF: }3\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}2 & 0 & -1\\0 & 2 & -1\\-1 & -1 & 1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & - \frac{1}{2}\\0 & 1 & - \frac{1}{2}\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}\frac{1}{2}\\\frac{1}{2}\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }1\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}0 & 0 & -1\\0 & 0 & -1\\-1 & -1 & -1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\1\\0\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }0\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}-1 & 0 & -1\\0 & -1 & -1\\-1 & -1 & -2\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\-1\\1\end{matrix}\right]\right]$

Out[32]:

$\left[\begin{array}{ccc}-1 & -1 & 1\\-1 & 1 & 1\\1 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 3\end{array}\right]\left[\begin{matrix}- \frac{1}{3} & - \frac{1}{3} & \frac{1}{3}\\- \frac{1}{2} & \frac{1}{2} & 0\\\frac{1}{6} & \frac{1}{6} & \frac{1}{3}\end{matrix}\right]$

In [33]:

# To fix the order, you can use the following trick
order = [2, 1, 0] # specify the order of singular values as it is
newP = pdp.P.select_cols(*order)
diag_entries = pdp.D.diagonal()
newD = Matrix.diag(*[diag_entries[i] for i in order])
newP, newD

# To fix the order, you can use the following trick order = [2, 1, 0] # specify the order of singular values as it is newP = pdp.P.select_cols(*order) diag_entries = pdp.D.diagonal() newD = Matrix.diag(*[diag_entries[i] for i in order]) newP, newD

Out[33]:

$\displaystyle \left( \left[\begin{array}{ccc}1 & -1 & -1\\1 & 1 & -1\\2 & 0 & 1\end{array}\right], \ \left[\begin{array}{ccc}3 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{array}\right]\right)$

In [34]:

newP @ newD @ newP.inv()

newP @ newD @ newP.inv()

Out[34]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 1\\1 & 1 & 2\end{array}\right]$

In [35]:

pdpt = A.orthogonally_diagonalize(verbosity=1, sort=True)
pdpt

pdpt = A.orthogonally_diagonalize(verbosity=1, sort=True) pdpt

Check if matrix is symmetric: True
Characteristic Polynomial

$\displaystyle x \left(x - 3\right) \left(x - 1\right)$

$\displaystyle \text{Before RREF: }3\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}2 & 0 & -1\\0 & 2 & -1\\-1 & -1 & 1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & - \frac{1}{2}\\0 & 1 & - \frac{1}{2}\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}\frac{1}{2}\\\frac{1}{2}\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }1\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}0 & 0 & -1\\0 & 0 & -1\\-1 & -1 & -1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\1\\0\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }0\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}-1 & 0 & -1\\0 & -1 & -1\\-1 & -1 & -2\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 1\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\-1\\1\end{matrix}\right]\right]$

Out[35]:

$\left[\begin{array}{ccc}- \frac{\sqrt{3}}{3} & - \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6}\\- \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6}\\\frac{\sqrt{3}}{3} & 0 & \frac{\sqrt{6}}{3}\end{array}\right]\left[\begin{array}{ccc}0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 3\end{array}\right]\left[\begin{matrix}- \frac{\sqrt{3}}{3} & - \frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3}\\- \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\\\frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{3}\end{matrix}\right]$

In [36]:

# Similar to diagonalization, the singular values are 
# not guaranteed to be in the correct order
order = [2, 1, 0] # specify the order of singular values as it is
newP = pdpt.P.select_cols(*order)
diag_entries = pdpt.D.diagonal()
newD = Matrix.diag(*[diag_entries[i] for i in order])
newP, newD

# Similar to diagonalization, the singular values are # not guaranteed to be in the correct order order = [2, 1, 0] # specify the order of singular values as it is newP = pdpt.P.select_cols(*order) diag_entries = pdpt.D.diagonal() newD = Matrix.diag(*[diag_entries[i] for i in order]) newP, newD

Out[36]:

$\displaystyle \left( \left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & - \frac{\sqrt{2}}{2} & - \frac{\sqrt{3}}{3}\\\frac{\sqrt{6}}{6} & \frac{\sqrt{2}}{2} & - \frac{\sqrt{3}}{3}\\\frac{\sqrt{6}}{3} & 0 & \frac{\sqrt{3}}{3}\end{array}\right], \ \left[\begin{array}{ccc}3 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{array}\right]\right)$

In [37]:

newP @ newD @ newP.T

newP @ newD @ newP.T

Out[37]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 1\\1 & 1 & 2\end{array}\right]$

Question 6¶

Let $\mathbf{A} = \begin{pmatrix} -18 & 13 & -4 & 4 \\ 2 & 19 & -4 & 12 \\ -14 & 11 & -12 & 8 \\ -2 & 21 & 4 & 8 \end{pmatrix}$.

(a)¶

Find a SVD of $\mathbf{A}$.

In [38]:

A = Matrix.from_str("-18 13 -4 4; 2 19 -4 12; -14 11 -12 8; -2 21 4 8")
A

A = Matrix.from_str("-18 13 -4 4; 2 19 -4 12; -14 11 -12 8; -2 21 4 8") A

Out[38]:

$\displaystyle \left[\begin{array}{cccc}-18 & 13 & -4 & 4\\2 & 19 & -4 & 12\\-14 & 11 & -12 & 8\\-2 & 21 & 4 & 8\end{array}\right]$

In [39]:

A.singular_value_decomposition(verbosity=1)

A.singular_value_decomposition(verbosity=1)

A^T A

$\displaystyle \left[\begin{array}{cccc}528 & -392 & 224 & -176\\-392 & 1092 & -176 & 536\\224 & -176 & 192 & -128\\-176 & 536 & -128 & 288\end{array}\right]$

Check if matrix is symmetric: True
Characteristic Polynomial

$\displaystyle x \left(x - 1600\right) \left(x - 400\right) \left(x - 100\right)$

$\displaystyle \text{Before RREF: }1600\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}1072 & 392 & -224 & 176\\392 & 508 & 176 & -536\\-224 & 176 & 1408 & 128\\176 & -536 & 128 & 1312\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 0 & 0 & 1\\0 & 1 & 0 & -2\\0 & 0 & 1 & \frac{1}{2}\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}-1\\2\\- \frac{1}{2}\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }400\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}-128 & 392 & -224 & 176\\392 & -692 & 176 & -536\\-224 & 176 & 208 & 128\\176 & -536 & 128 & 112\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 0 & 0 & -4\\0 & 1 & 0 & -2\\0 & 0 & 1 & -2\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}4\\2\\2\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }100\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}-428 & 392 & -224 & 176\\392 & -992 & 176 & -536\\-224 & 176 & -92 & 128\\176 & -536 & 128 & -188\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 0 & 0 & -1\\0 & 1 & 0 & \frac{1}{2}\\0 & 0 & 1 & 2\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}1\\- \frac{1}{2}\\-2\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }0\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}-528 & 392 & -224 & 176\\392 & -1092 & 176 & -536\\-224 & 176 & -192 & 128\\176 & -536 & 128 & -288\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 0 & 0 & \frac{1}{4}\\0 & 1 & 0 & \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2}\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}- \frac{1}{4}\\- \frac{1}{2}\\\frac{1}{2}\\1\end{matrix}\right]\right]$

$\displaystyle u_1 = (1/40)A\left[\begin{matrix}- \frac{2}{5}\\\frac{4}{5}\\- \frac{1}{5}\\\frac{2}{5}\end{matrix}\right] = \left[\begin{array}{c}\frac{1}{2}\\\frac{1}{2}\\\frac{1}{2}\\\frac{1}{2}\end{array}\right]$

$\displaystyle u_2 = (1/20)A\left[\begin{matrix}\frac{4}{5}\\\frac{2}{5}\\\frac{2}{5}\\\frac{1}{5}\end{matrix}\right] = \left[\begin{array}{c}- \frac{1}{2}\\\frac{1}{2}\\- \frac{1}{2}\\\frac{1}{2}\end{array}\right]$

$\displaystyle u_3 = (1/10)A\left[\begin{matrix}\frac{2}{5}\\- \frac{1}{5}\\- \frac{4}{5}\\\frac{2}{5}\end{matrix}\right] = \left[\begin{array}{c}- \frac{1}{2}\\\frac{1}{2}\\\frac{1}{2}\\- \frac{1}{2}\end{array}\right]$

Extending U with its orthogonal complement.
Before RREF: [self]

$\displaystyle \left[\begin{matrix}\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & - \frac{1}{2}\end{matrix}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 0 & 0 & 1\\0 & 1 & 0 & 1\\0 & 0 & 1 & -1\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

$\displaystyle v_1 = \left[\begin{array}{c}-1\\-1\\1\\1\end{array}\right]$

Out[39]:

$\left[\begin{array}{cccc}\frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & - \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & \frac{1}{2}\end{array}\right]\left[\begin{array}{cccc}40 & 0 & 0 & 0\\0 & 20 & 0 & 0\\0 & 0 & 10 & 0\\0 & 0 & 0 & 0\end{array}\right]\left[\begin{array}{cccc}- \frac{2}{5} & \frac{4}{5} & - \frac{1}{5} & \frac{2}{5}\\\frac{4}{5} & \frac{2}{5} & \frac{2}{5} & \frac{1}{5}\\\frac{2}{5} & - \frac{1}{5} & - \frac{4}{5} & \frac{2}{5}\\- \frac{1}{5} & - \frac{2}{5} & \frac{2}{5} & \frac{4}{5}\end{array}\right]$