Math599 2022S¶

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LA exam¶

Please find the ipynb and related files for this exam through the link below.
https://hackmd.io/@jephianlin/HyTpzFAG5

Name:

Student ID #:

Please read the instructions carefully:

  1. Write your name and Student ID # first.
  2. Write down your answers on the exam paper; no need to write down the code.
  3. You are allowed to use the internet, but you are NOT allowed to communicate with others in any form.
  4. Do NOT use your cell phone, use the computer instead.
  5. Different problems might use same variable names. Make sure you use the right one to answer the problem.
  6. If the answer is too long, write two digits after the decimal point.
In [1]:
import numpy as np
import matplotlib.pyplot as plt
Problem 1 [2pt]¶

Let

ts = np.linspace(5,10,101)
xs = np.sin(ts)
ys = np.log(ts)
X = np.vstack([xs,ys]).T

Let $\bv_0,\ldots,\bv_{100}$ be the rows of X .
Find the center

$$ \frac{1}{101}\sum_{i = 0}^{100}\bv_i. $$

Your answer:

In [2]:
ts = np.linspace(5,10,101)
xs = np.sin(ts)
ys = np.log(ts)
X = np.vstack([xs,ys]).T

X.mean(axis=0)
Out[2]:
array([0.21483686, 1.99533487])
Problem 2 [2pt]¶

Let

ts = np.linspace(5,10,101)
xs = np.sin(ts)
ys = np.log(ts)

Let $x_0,\ldots, x_{100}$ and $y_0,\ldots,y_{100}$ be the values in xs and ys , respectively.
Consider the points $(x_i,y_i)$ for $i = 0,\ldots, 100$.
How many of them satisfies $x^2 + 2y < 4$?

Your answer:

In [3]:
ts = np.linspace(5,10,101)
xs = np.sin(ts)
ys = np.log(ts)

np.sum(xs**2 + 2*ys < 4)
Out[3]:
28
Problem 3 [2pt]¶

Let

xs = np.linspace(5,10,101)
ys = 3.1*xs + 1.9 + 0.7*np.cos(xs)

Let $x_0,\ldots, x_{100}$ and $y_0,\ldots,y_{100}$ be the values in xs and ys , respectively.
Let $f(x) = 3x + 2$.
Find the mean square error between $y$ and $f(x)$, which is defined as
$$\frac{1}{101}\sum_{i=0}^{100}(f(x_i) - y_i)^2.$$

Your answer:

In [4]:
xs = np.linspace(5,10,101)
ys = 3.1*xs + 1.9 + 0.7*np.cos(xs)

((3*xs + 2 - ys) ** 2).mean()
Out[4]:
0.658567185219327
Problem 4 [2pt]¶

Let

ts = np.linspace(0, np.pi, 181)
xs = 3 * np.cos(ts)
ys = 2 * np.sin(ts)

Let $x_0,\ldots, x_{180}$ and $y_0,\ldots,y_{180}$ be the values in xs and ys , respectively.
Consider the points $(x_i,y_i)$ for $i = 0,\ldots, 180$.
For which $i$, the point $(x_i,y_i)$ is the closest to the point $(3,4)$?

Your answer:

In [5]:
ts = np.linspace(0, np.pi, 181)
xs = 3 * np.cos(ts)
ys = 2 * np.sin(ts)

np.argmin((xs - 3)**2 + (ys - 4)**2)
Out[5]:
53
Problem 5 [2pt]¶

Let

$$ A = \begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{bmatrix} \text{ and } \bb = \begin{bmatrix} 3 \\ 3 \\ 4 \end{bmatrix}. $$

Find $\bx\in\mathbb{R}^2$ such that $\|A\bx - \bb\|$ is minimized.

Your answer:

In [6]:
A = np.array([
    [1,1],
    [1,2],
    [1,3]
])
b = np.array([3,3,4])

ATAinv = np.linalg.inv(A.T.dot(A))
x = ATAinv.dot(A.T).dot(b)
x
Out[6]:
array([2.33333333, 0.5       ])
Problem 6 [2pt]¶

Let

$$ A = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{bmatrix} \text{ and } \bv = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}. $$

Is $\bv$ an eigenvector of $A$?
If yes, what is the corresponding eigenvalue?

Your answer:

In [7]:
A = np.array([
    [3,1,1],
    [1,3,1],
    [1,1,3]
])
v = np.array([1,-1,0])

print(A.dot(v))
print(v)
# Observe that A v = 2 v.  
# Yes, v is the eigenvector of A and the corresponding eigenvalue is 2.

[ 2 -2  0]
[ 1 -1  0]
Problem 7 [2pt]¶

Let

Q = np.array([[1,1,1], 
              [-1,1,1], 
              [0,-2,1]])
Q = Q / np.linalg.norm(Q, axis=0)

Let $\bu_0, \bu_1, \bu_2$ be the columns of Q .
Find a matrix $A$ such that
$A\bu_0 = 3\bu_0$
$A\bu_1 = 4\bu_1$
$A\bu_2 = 5\bu_2$.

Your answer:

In [8]:
Q = np.array([[1,1,1], 
              [-1,1,1], 
              [0,-2,1]])
Q = Q / np.linalg.norm(Q, axis=0)

D = np.diag([3,4,5])
A = Q.dot(D).dot(Q.T)
A
Out[8]:
array([[3.83333333, 0.83333333, 0.33333333],
       [0.83333333, 3.83333333, 0.33333333],
       [0.33333333, 0.33333333, 4.33333333]])
Problem 8 [2pt]¶

Let

Q = np.array([[1,2,2], 
              [-2,1,1], 
              [0,-5,1]])
Q = Q / np.linalg.norm(Q, axis=0)
D = np.diag([1,1,0])
A = Q.dot(D).dot(Q.T)

Let $A$ be the same as A .
What does the action $\bv \mapsto A\bv$ do?
If it is a projection or a reflection, make sure you describe the corresponding line/plane.

Your answer:

In [ ]:
# Since the eigenvalues are 1, 1, 0,  
# the action is the projection onto the plane spanned by the 0-th and the 1-st columns of Q.
Problem 9 [2pt]¶

Let

$$ A = \begin{bmatrix} 1 & 3 \\ 1 & 4 \\ 1 & 5 \end{bmatrix}. $$

Find the minimum value of $\bx\trans A\trans A\bx$ over all vector $\bx\in\mathbb{R}^2$ of length $1$.

Your answer:

In [9]:
A = np.array([
    [1,3],
    [1,4],
    [1,5]
])
ATA = A.T.dot(A)

vals, vecs = np.linalg.eigh(ATA)
vals.min()
Out[9]:
0.11345039608247687
Problem 10 [2pt]¶

Go to the link provided on the top of this file, and download the file hidden_text.csv . Let

arr = np.genfromtxt('hidden_text.csv', delimiter=',')
m,n = arr.shape
U,s,Vh = np.linalg.svd(arr)

Let $\sigma$'s be the values in s, $\bu_i$'s the columns of U, and $\bv_i$'s the columns of Vh.T .
Compute
$$ B = \sum_{\sigma < 300}\sigma_i\bu_i\bv_i\trans $$ and use plt.imshow(..., cmap='Greys_r') to visualize the matrix $B$.
What text do you see there?

Your answer:

In [ ]:
arr = np.genfromtxt('hidden_text.csv', delimiter=',')
m,n = arr.shape
U,s,Vh = np.linalg.svd(arr)

plt.plot(s) ### check the values in s --> lots of values are above 400
In [ ]:
s[s > 300] = 0
Sigma = np.zeros((m,n))
Sigma[np.arange(m), np.arange(m)] = s # m is the smaller one
new_arr = U.dot(Sigma).dot(Vh)
plt.imshow(new_arr, cmap='Greys_r') ### so the answer is LOVE
Problem 11 [extra 2pt]¶

Run code cell 1 and code cell 2.
It is known that f is a linear function from $\mathbb{R}^3$ to $\mathbb{R}^3$.
You may modify the line

v = f([1, 2, 3])

in code cell 2 to find out the output of f with different inputs.
Find a matrix $A$ such that f$(\bx) = A\bx$ for all $\bx\in\mathbb{R}^3$.

In [ ]:
# code cell 1
bit_encoding = {0:'0',1:'1',2:'2',3:'3',4:'4',5:'5',6:'6',7:'7',8:'8',9:'9',10:'A',11:'B',12:'C',13:'D',14:'E',15:'F',16:'G',17:'H',18:'I',19:'J',20:'K',21:'L',22:'M',23:'N',24:'O',25:'P',26:'Q',27:'R',28:'S',29:'T',30:'U',31:'V',32:'W',33:'X',34:'Y',35:'Z',36:'a',37:'b',38:'c',39:'d',40:'e',41:'f',42:'g',43:'h',44:'i',45:'j',46:'k',47:'l',48:'m',49:'n',50:'o',51:'p',52:'q',53:'r',54:'s',55:'t',56:'u',57:'v',58:'w',59:'x',60:'y',61:'z',62:'-',63:'%'}
def get_parameter_encoding(data):
    data = bytes(data, encoding='UTF8')
    data_len = len(data)
    data += b'\0\0\0'
    encode_len = (data_len * 8) // 6 + (data_len % 3 > 0)
    encode = bytearray(encode_len)
    index = 0
    read_mode = [((int('00111111', 2), 0), (int('00000000', 2), 0), 0), ((int('11000000', 2), 6), (int('00001111', 2), 2), 1), ((int('11110000', 2), 4), (int('00000011', 2), 4), 1), ((int('11111100', 2), 2), (int('00000000', 2), 0), 1)]
    for i in range(encode_len):
        r_mod = read_mode[i % 4]
        lbits = data[index] & r_mod[0][0]
        hbits = data[index + 1] & r_mod[1][0]
        encode[i] = ord(bit_encoding[(lbits >> r_mod[0][1]) + (hbits << r_mod[1][1])])
        index += r_mod[2]
    return encode.decode(encoding='UTF8')
In [ ]:
# code cell 2
import requests

result = requests.post('https://python-online-runner.herokuapp.com/api/pyrunner', {
    'define': 'AGMPc1YPe4cSobYEA028W02Uia7Bw1IFW4cSof08W028oL6Tr9dRWiLUhe7BWe7BWWNNA0',
    'execute': get_parameter_encoding('''

v = f([1, 2, 3])
print(v)
''')
})

print(result.text)

Your answer:

In [ ]:
# Use the code to get 
# f(1,0,0) --> (0,0,1)
# f(0,1,0) --> (1,0,0)
# f(0,0,1) --> (1,1,0)
# Therefore, A can be 
# [0, 1, 1]
# [0, 0, 1]
# [1, 0, 0]

Exam ends here.
Total point = 20 (+2)

Your score: