## Description

## Question 1.

For this question, we will work in the floating-point system F(β = 10, p = 3, L = −10, U = 10).

Consider the system Ax = b where

A =

0.03 21.2

2.18 −0.30

, b =

21.5

21.5

## Part (a) – 2pt

Solve the system by hand, exactly, using infinite precision arithmetic. That is, the addition, multiplication, division,

and other operations that you do will be exact (rather than the floating-point version). Do not use pivoting.

Include your step-by-step solution in your PDF writeup. Your final result can be rounded to 5 decimal places, but do

not round any intermediate computations since we are using infinite precision arithmetic.

## Part (b) – 2pt

Solve the system using floating-point operations and the rounding rule is chopping. In other words, replace

every operation you used in part (a) with the floating-point operation in the system F(β = 10, p = 3, L = −10, U = 10).

Do not use pivoting.

Include your step-by-step solution in your PDF writeup.

## Part (c) – 3pt

Solve this similar system using floating-point operations, but with partial pivoting.

Which result is closer to the true result that you obtained in Part (a)?

Include your step-by-step solution in your PDF writeup.

1

##
Question 2.

Part (a) [2 pt]

Prove that ||A||∞ can be computed by computing the maximum row sum:

||A||∞ = max

i

Xn

j=1

|aij |

## Part (b) [2 pt]

Suppose that A and B are two n × n matrices, and both are well-conditioned. Is A(B−1

) also well-conditioned? Why

or why not? Include your answer and justifications in your pdf write-up. Be specific.

## Part (c) [4 pt]

Describe an efficient algorithm to compute d

T BT A−1Bd

Where:

• A is an invertible n × n matrix,

• B is an n × n matrix, and

• d is an n × 1 vectors

Be clear and specific. Include your strategy in your pdf write-up.

##
Question 3

Part (a) – 2 pt

Suppose that A is a symmetric positive definite matrix. Show that the function

||x||A = (x

T Ax)

1

2

on a vector x satisfies the three properties of a vector norm.

Include your solution in your pdf writeup.

## Part (b) – 6 pt

Complete the function cholesky_factorize that returns the Cholesky factorization of a matrix, according to its

docstring.

def cholesky_factorize(A):

“””Return the Cholesky Factorization L of A, where

* A is an nxn symmetric, positive definite matrix

* L is lower triangular, with positive diagonal entries

* $A = LLˆT$

>>> M = np.array([[8., 3., 2.],

[3., 5., 1.],

[2., 1., 3.]])

>>> L = cholesky_factorize(M)

>>> np.matmul(L, L.T)

array([[8., 3., 2.],

2

[3., 5., 1.],

[2., 1., 3.]])

“””

##
Question 4

Part (a) – 4 pts

Complete the function solve_rank_one_update that solves the system (A − uvT

)x = b, assuming that the factorization A = LU has already been done for you. You are welcome to add any helper functions that you wish, including

functions that you wrote in homeworks 3 and 4. Just make sure that you include the helper functions in your python

script submission.

def solve_rank_one_update(L, U, b, u, v):

“””Return the solution x to the system (A – u vˆT)x = b, where

A = LU, using the approach we derived in class using

the Sherman Morrison formula. You may assume that

the LU factorization of A has already been computed for you, and

that the parameters of the function have:

* L is an invertible nxn lower triangular matrix

* U is an invertible nxn upper triangular matrix

* b is a vector of size n

* u and b are also vectors of size n

>>> A = np.array([[2., 0., 1.],

[1., 1., 0.],

[2., 1., 2.]])

>>> L, U = lu_factorize(A) # from homework 3

>>> L

array([[1. , 0. , 0. ],

[0.5, 1. , 0. ],

[1. , 1. , 1. ]])

>>> U

array([[ 2. , 0. , 1. ],

[ 0. , 1. , -0.5],

[ 0. , 0. , 1.5]])

>>> b = np.array([1., 1., 0.])

>>> u = np.array([1., 0., 0.])

>>> v = np.array([0., 2., 0.])

>>> x = solve_rank_one_update(L, U, b, u, v)

>>> x

array([1. , 0. , -1.])

>>> np.matmul((A – np.outer(u, v)), x)

array([1. , 1. , 0.])

“””

## Part (b) – 2 pt

Explain why using solve_rank_one_update does not give us accurate results in the below example:

def run_example():

A = np.array([[2., 0., 1.],

[1., 1., 0.],

[1., 1., 1.]])

L = np.array([[1., 0., 0.],

[0.5, 1., 0.],

3

[0.5, 1., 1.]])

U = np.array([[2., 0., 1.],

[0., 1., -0.5],

[0., 0., 1.]])

b = np.array([1, 1, -1])

u = np.array([0, 0, 0.9999999999999999])

v = np.array([0, 0, 0.9999999999999999])

x = solve_rank_one_update(L, U, b, u, v)

print(np.matmul((A – np.outer(u, v)), x) – b)

Include your solution in your PDF writeup.

## Part (c) – 4 pt

A rank 1 matrix has the form xyT where x and y are column vectors. Suppose A and B are non-sigular matrices.

Show that A − B is rank 1 if and only if A−1 − B−1

is also rank 1.

Include your solution in your pdf writeup.

Hint: Use the Sherman-Morrison formula. This question is not supposed to be easy, so leave aside time to think!

##
Question 5.

Part (a) – 3 pt

Write a function householder_v that returns the vector v that defines the Householder transform

H = I − 2

vvT

v

T v

that eliminates all but the first element of a vector a. You may use the function np.linalg.norm.

def householder_v(a):

“””Return the vector $v$ that defines the Householder Transform

H = I – 2 np.matmul(v, v.T) / np.matmul(v.T, v)

that will eliminate all but the first element of the

input vector a. Choose the $v$ that does not result in

cancellation.

Do not modify the vector `a`.

Example:

>>> a = np.array([2., 1., 2.])

>>> householder_v(a)

array([5., 1., 2.])

>>> a

array([2., 1., 2.])

“””

## Part (b) – 2 pt

Show that a Householder Transformation H is orthogonal. Include your solution in your PDF writeup.

4

## Part (c) – 2 pt

Write a function apply_householder that applies the Householder transform defined by a vector v to a vector u.

You should not compute the Householder transform matrix H. You should only need to compute vector-vector dot

products and vector-scalar multiplications.

def apply_householder(v, u):

“””Return the result of the Householder transformation defined

by the vector $v$ applied to the vector $u$. You should not

compute the Householder matrix H directly.

Example:

>>> apply_householder(np.array([5., 1., 2.]), np.array([2., 1., 2.]))

array([-3., 0., 0.])

>>> apply_householder(np.array([5., 1., 2.]), np.array([2., 3., 4.]))

array([-5. , 1.6, 1.2])

“””

## Part (d) – 3 pt

Write a function apply_householder_matrix that applies the Householder transform defined by a vector v to all the

columns of a matrix U. You should not compute the Householder transform matrix H.

Do not use for loops. Instead, you may find the numpy function np.outer useful.

def apply_householder_matrix(v, U):

“””Return the result of the Householder transformation defined

by the vector $v$ applied to all the vectors in the matrix U.

You should not compute the Householder matrix H directly.

Example:

>>> v = np.array([5., 1., 2.])

>>> U = np.array([[2., 2.],

[1., 3.],

[2., 4.]])

>>> apply_householder_matrix(v, U)

array([[-3. , -5. ],

[ 0. , 1.6],

[ 0. , 1.2]])

“””

## Part (e) – 3 pt

Write a function solve_qr_householder that takes an m × n matrix A and a vector b, and solves the linear least

squares problem Ax ≈ b using Householder QR Factorization. You may use np.linalg.solve to solve any square

system of the form Ax = b that you produce.

You should use the helper function qr_householder that takes a matrix A and a vector b and performs the Householder

QR Factorization using the functions you wrote in parts (b-d).

def qr_householder(A, b):

“””Return the matrix [R O]ˆT, and vector [c1 c2]ˆT equivalent

to the system $Ax \approx b$. This algorithm is similar to

Algorithm 3.1 in the textbook.

“””

for k in range(A.shape[1]):

5

v = householder_v(A[k:, k])

if np.linalg.norm(v) != 0:

A[k:, k:] = apply_householder_matrix(v, A[k:, k:])

b[k:] = apply_householder(v, b[k:])

# now, A is upper-triangular

return A, b

def solve_qr_householder(A, b):

“””

Return the solution x to the linear least squares problem

$$Ax \approx b$$ using Householder QR decomposition.

Where A is an (m x n) matrix, with m > n, rank(A) = n, and

b is a vector of size (m)

“””

## Question 6

For the next few questions, we will use the MNIST dataset for digit recognition Download the files mnist_images.npy

and mnist_labels.npy from the course website, and place them into the same folder as your ipynb file.

The code below loads the data, splits it into “train” and “test” sets, and plots the a subset of the data. We will

use train_images and train_labels to set up Linear Least Squares problems. We will use test_images and

test_labels to test the models that we build.

mnist_images = np.load(“mnist_images.npy”)

test_images = mnist_images[4500:] # 500 images

train_images = mnist_images[:4500] # 4500 images

mnist_labels = np.load(“mnist_labels.npy”)

test_labels = mnist_labels[4500:]

train_labels = mnist_labels[:4500]

def plot_mnist(remove_border=False):

“”” Plot the first 40 data points in the MNIST train_images. “””

import matplotlib.pyplot as plt

plt.figure(figsize=(10, 5))

for i in range(4 * 10):

plt.subplot(4, 10, i+1)

if remove_border:

plt.imshow(train_images[i,4:24,4:24])

else:

plt.imshow(train_images[i])# plot_mnist() # please comment out this line before submission

Remember to comment out any plotting related code before you submit your assignment!

# plot_mnist()

## Part (a) – 1 pt

How many examples of each digit are in train_images? You should use the information in train_labels.

Some of the code in the later part of this question might be helpful. You might also find the sum function helpful.

Save your result in the array mnist_digits, where mnist_digits[0] should contain the number of digit 0 in

train_images, mnist_digits[1] should contain the number of digit 1 in train_images, etc.

mnist_digits = []

6

## Part (b) – 2 pt

We will build a rudimentary model to predict whether a digit is the digit 0. Our features will be the intensity at each

pixel. There are 28 * 28 = 784 pixels in each image. However, in order to obtain a matrix A that is of full rank, we

will ignore the pixels along the border. That is, we will only use 400 pixels in the center of the image.

Look at a few of the MNIST images using the plot_mnist function written for you. Why would our matrix A not be

full rank if we use all 784 pixels in our model?

If this question doesn’t make sense yet, you might want to come back to it after completing the rest of this question.

Include your solution in your PDF writeup.

# Plot the MNIST images, with only the center pixels that we will use

# in our digit classification model.

#plot_mnist(True) # please comment out this line before submission

## Part (c) – 1 pt

We will now build a rudimentary model to predict whether a digit is the digit 0. To obtain a matrix of full rank, we

will use the 400 pixels at the center of each image as features. Our target will be whether our digit is 0.

In short, the model we will build looks like this:

x1p1 + x2p2 + … + x400p400 = y

Where pi

is the pixel intensity at pixel i (the ordering of the pixel’s doesn’t actually matter), and the value of y

determines whether our digit is a 0 or not.

We will solve for the coefficients xi by solving the linear least squares problem Ax ≈ b, where A is constructed using

the pixel intensities of the images in train_images, and y is constructed using the labels for those images. For

convenience, we will set y = 1 for images of the digit 0, and y = 0 for other digits.

We should stress that in real machine learning courses, you will learn that this is not the proper way

to build a digit detector. However, digit detection is quite fun, so we might as well use to tools that we have to

try and solve the problem.

The code below obtains the matrices A and the vector b of our least squares problem, where A is a m × n matrix and

b is a vector of length m.

What is the value of m and n? Save the values in the variables mnist_m and mnist_n.

A = train_images[:, 4:24, 4:24].reshape([-1, 20*20])

b = (train_labels == 0).astype(np.float32)

mnist_m = None

mnist_n = None

## Part (d) – 1 pt

Use the Householder QR decomposition method to solve the system. Save the result in the variable mnist_x. Save

the norm of the residuals of this solution in the variable mnist_r.

mnist_x = None

mnist_r = None

7

## Part (e) – 1 pt

Consider test_images[0]. Is this image of the digit 0? Set the value of test_image_0 to either True or False

depending on your result.

Let p be the vector containing the values of the 400 center pixels in test_image[0]. The features are extracted for

you in the variabel p. Use the solution mnist_x to estimate the target y for the vector p. Save the (float) value of the

predicted value of y in the variable test_image_0_y.

# import matplotlib.pyplot as plt # Please comment before submitting

# plt.imshow(test_images[0]) # Please comment before submitting

p = test_images[0, 4:24, 4:24].reshape([-1, 20*20])

test_image_0 = None

test_image_0_y = None

## Part (f) – 2 pt

Write code to predict the value of y for every image in test_images. Save your result in test_image_y.

Do not use a loop.

test_image_y = None

## Part (g) – 1 pt

We will want to turn the continuous estimates of y into discrete predictions about whether an image is of the digit 0.

We will do this by selecting a cutoff. That is, we will predict that a test image is of the digit 0 if the prediction y for

that digit is at least 0.5.

Create a numpy array test_image_pred with test_image_pred[i] == 1 if test_image[i] is of the digit 0, and

test_image_pred[i] == 0 otherwise. Then, run the code to compute the test_accuracy, or the portion of the

times that our prediction matches the actual label.

## HINT:

You might find the code in Part(c) helpful.

(This is somewhat of an arbitrary cutoff. You will learn the proper way to do this prediction problem in a machine

learning course like CSC311.)

test_image_pred = None

# test_accuracy = sum(test_image_pred == (test_labels == 0).astype(float)) / len(test_labels)

# print(test_accuracy)

## Part (h) – 4 pt

So far, we built a linear least squares model that determines whether an image is of the digit 0. Let’s go a step

further, and build such a model for every digit!

Complete the function mnist_classifiers that uses train_images and train_labels, and the functions you wrote

in Q3 to build a linear least squares model for every digit. The function should return a matrix xs of shape 10 × 400,

with xs[0] == mnist_x1 from earlier.

Make sure to comment out any code you use to test mnist_classifier, or your code might not be

testable.

This part of the code will be graded by your TA.

def mnist_classifiers():

“””Return the coefficients for linear least squares models for every digit.

Example:

8

>>> xs = mnist_classifiers()

>>> np.all(xs[0] == mnist_x1)

True

“””

# you can use train_images and train_labels here, and make

# whatever edits to this function as you wish.

A = train_images[:, 4:24, 4:24].reshape([-1, 20*20])

xs = []

# …

return np.stack(xs)

Just for fun. . .

The code below makes predictions based on the result of your mnist_classifier. That is, for every test image, the

code runs all 10 models to see whether the test image contains each of the 10 digits. We make a discrete prediction

about which digit the image contains by looking at which model yields the largest value of y for the image.

The code then compares the result against the actual labels, computes the accuracy measure: the fraction of predictions

that is correct. Just for fun, look at the prediction accuracy of our model, but please comment any code you write

before submitting your assignment.

Again, in machine learning and statistics courses you will learn ways to classifying digits that are better and more

principled. You’ll achieve a much better test accuracy than what we have here.

def prediction_accuracy(xs):

“””Return the prediction

“””

testA = test_images[:, 4:24, 4:24].reshape([-1, 20*20])

ys = np.matmul(testA, xs.T)

pred = np.argmax(ys, axis=1)

return sum(pred == test_labels) / len(test_labels)

9