# MATLAB PROJECT 4 SOLVED

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MATLAB PROJECT 4
Please read the Instructions located on the Assignments page prior
to working on the Project.
BEGIN with creating Live Script Project4.
Note: All exercises in this project will be completed in the Live Script using the Live Editor.
Each exercise has to begin with the line
Exercise#
You should also mark down the parts such as (a), (b), (c), and etc. This makes grading easier.
Important: we use the default format short for the numbers in all exercises unless it is
specified otherwise. If format long or format rat has been used, please make sure to
return to the default format in the next exercise.
Part I. Eigenvalues, Eigenvectors & Diagonalization
Exercise 1 (5 points) Difficulty: Hard
In this exercise, you will work with the eigenvalues of a given n n × matrix A. First, you will
find all eigenvalues and, then, you will take distinct eigenvalues and find orthonormal bases for
the corresponding eigenspaces and the dimensions of the eigenspaces. Next, you will check if
A is diagonalizable by applying the general theory. If a matrix A is diagonalizable, the output
has to contain an invertible matrix P and the diagonal matrix D, such that, 1 A PDP− = , or,
equivalently, AP PD = and P is invertible. Next, for diagonalizable matrices, you will run a
built-in MATLAB function, which also performs diagonalization, and you will compare its
outputs with the outputs P and D of your function.
**Create a function in MATLAB that begins with
function [P,D]=eigen(A)
format compact
[~,n]=size(A);
Your function [P,D]=eigen(A)will have a set of commands which generates the following
outputs for an n n × matrix A. Each output has to be supplied with a corresponding message –
you could use the commands disp and fprintf.
Part 1. Output the vector L of eigenvalues of the matrix.
To do that, you will go through several steps. Please suppress all intermediate outputs and
display only the final output vector L.
**Find a row vector L of all eigenvalues, where each eigenvalue repeats as many times as its
multiplicity. A basic MATLAB command for this part L=eig(A) returns a column vector of all
eigenvalues of A, and we use the function transpose() to get a row vector.
**You will also need to sort the entries of L – use the MATLAB command sort() to output
the entries of L in ascending order.
2
To output vector L correctly, you will go through two more steps:
**In your code, you have to ensure that the multiple eigenvalues show as the same number.
You will use closetozeroroundoff function with the parameter p = 7 to compare the
eigenvalues and assign the same value to the ones that are within 10^(-7) tolerance of each
other. To perform this task, you can go through the sequence of the sorted eigenvalues and, if
there is a set of eigenvalues that are within 10^(-7) from each other, you will assign the first
eigenvalue that comes across to all of them in that set.
Note: please work with the eigenvalues that are stored in MATLAB – we do not round off any
eigenvalues on this step.
**A singular matrix A has a zero eigenvalue; however, when running your code, a zero
eigenvalue may not show as a zero due to round-off errors in MATLAB. If a matrix A is not
invertible, you will run the function closetozeroroundoff with p = 7 on the vector of the
eigenvalues to ensure that the zero eigenvalues show as a zero. To check whether a matrix is
not invertible, please use a MATLAB command rank().
After performing all the tasks outlined above, you will display your final output L – a row
vector of the sorted eigenvalues of A where all multiple eigenvalues equal to each other and
the zero eigenvalues (for singular matrices) show as a zero.
Part 2. Construct an orthonormal basis W for each eigenspace.
**Output a row vector M of all distinct eigenvalues (no repetitions are allowed). The
MATLAB command unique() can be used here.
For each distinct eigenvalue M(i), your function has to do the following:
**Find the multiplicity, m(i), of the eigenvalue M(i). Output it with the message:
fprintf(‘Eigenvalue %d has multiplicity %i\n’,M(i),m(i))
**Output an orthonormal basis W for the eigenspace for the eigenvalue M(i). Display it with
the message:
fprintf(‘A basis for eigenvalue lambda = %d is:\n’,M(i))
W
Hint: An appropriate command in MATLAB that creates an orthonormal basis for the null
space of a matrix is null().
**Calculate the dimension d(i) of W. Output it with the message:
fprintf(‘Dimension of eigenspace for lambda = %d is %i\n’,M(i),d(i))
Part 3. Construct a Diagonalization when possible.
(For help with this part, please review the material of Lecture 14)
**First, we determine if A is diagonalizable. You will work with the multiplicity of the
eigenvalue M(i), m(i), and the dimension of the corresponding eigenspace, d(i), for every i.
If A is not diagonalizable, output a corresponding message, assign the empty outputs
P=[];D=[];
and terminate the program.
If A is diagonalizable, output a corresponding message, and your function will continue with
constructing a diagonalization:
**Output and display a matrix P constructed from the combined bases W for the eigenspaces.
**Output and display a diagonal matrix D with the corresponding eigenvalues on its main
diagonal.
3
**Verify that both conditions hold: AP=PD and P is invertible. To verify the first condition,
you will need to use the function closetozeroroundoff with the parameter p = 7. To verify
the second condition, please use the command rank(). If both conditions hold, output a
message: ‘Great! I got a diagonalization!’ Otherwise, output a message: ‘Oops! I
got a bug in my code!’ and terminate the program.
Part 4. Compare your outputs with the corresponding ones of a MATLAB built-in function.
If the diagonalization is confirmed, your code will continue with the following task.
**There is a MATLAB built-in function that runs as
[U,V]=eig(A)
which, for a diagonalizable matrix A, generates a diagonal matrix V, with the eigenvalues of A
on its main diagonal, and an n n × invertible matrix U of the corresponding eigenvectors, such
that AU=UV. Place this function in your code and display the outputs U and V.
**Compare the output matrix U with the matrix P: write a set of commands that would check
on the following case:
The sets of columns of P and U are the same or match up to a scalar (-1), where the
order of the columns does not matter.
If it is the case, output a message ‘Sets of columns of P and U are the same or match
up to scalar (-1)’.
If it is not the case (which is possible), the output message could be ‘There is no specific
match among the columns of P and U’
Note: You will need to use the function closetozeroroundoff with p = 7 when comparing
the sets of columns of the matrices P and U.
**Verify that the matrices D and V have the same set of diagonal elements (the order does not
count either). If it is the case, output a message ‘The diagonal elements of D and V
match’. If it is not the case, output something like: ‘That cannot be true!’
Hint: To perform this task, you can “sort” the diagonal elements of V and compare them with
the vector L using the function closetozeroroundoff with p = 7.
This is the end of the function eigen.
**Print the functions eigen, closetozeroroundoff,jord in the Live Script.
(The function jord was created in Exercise 1 of Project 1.)
**Type in the Live Script
format
**Run the function [P,D]=eigen(A); on the following matrices:
%(a)
A=[3 3; 0 3]
%(b)
A=[4 0 0 0; 1 3 0 0; 0 -1 3 0; 0 -1 5 4]
%(c)
A=jord(5,5)
%(d)
A=diag([3, 3, 3, 2, 2, 1])
%(e)
A=magic(4)
4
%(f)
A=ones(4)
%(g)
A=magic(5)
%(h)
A=hilb(7)
%(k)
A=[5 8 -4;8 5 -4;-4 -4 -1]
Exercise 2 (3 points) Difficulty: Easy
In this exercise, we will construct an orthogonal diagonalization of an n n × symmetric matrix.
Theory: A square matrix A is called symmetric if T A A = .
A matrix A is said to be orthogonally diagonalizable if there exists an orthogonal matrix P
( 1 T P P − = ) and a diagonal matrix D, such that 1 A PDP− = , or equivalently, T A PDP = .
An n n × matrix A is orthogonally diagonalizable if and only if A is symmetric.
**Create a function in MATLAB
function []=symmetric(A)
**First, your function has to check whether A is symmetric. If not, output a message that the
matrix is not symmetric and terminate the program.
**If a matrix is symmetric, you will construct an orthogonal diagonalization using a built-in
MATLAB function
[P,D]=eig(A)
in your code. The outputs have to be an orthogonal matrix P of the eigenvectors of A and a
diagonal matrix D with the corresponding eigenvalues on its main diagonal. Place this function
in your code and display the outputs P and D.
**Next, you should verify in your code that you have got an orthogonal diagonalization, that
is, both conditions hold: AP=PD and P is an orthogonal matrix. If it is the case, output a
corresponding message. Otherwise, a message could be ‘What is wrong?!’
Note: you will need to use the function closetozeroroundoff with p = 7 in your code for
verification of each of the two conditions above.
(For the help with this exercise, please refer to Lecture 29.)
This is the end of the function symmetric.
**Print the functions symmetric and closetozeroroundoff in your Live Script.
**Run the function on the given matrices as indicated below:
%(a)
A=[2 -1 1;-1 2 -1;1 -1 2]
symmetric(A)
%(b)
A=[2 -1 1;-1 2 -2;1 -1 2]
symmetric(A)
%(c)
B=A*A’
symmetric(B)
5
%(d)
A=[3 1 1;1 3 1;1 1 3]
symmetric(A)
%(e)
A=[5 8 -4;8 5 -4;-4 -4 -1]
symmetric(A)
%(f)
A=[4 3 1 1; 3 4 1 1 ; 1 1 4 3; 1 1 3 4]
symmetric(A)
Part II. Orthogonal Projections & Least-Squares Solutions
Exercise 3 (5 points) Difficulty: Moderate
In this exercise, you will create a function proj(A,b)which will work with the projection of a
vector b onto the Column Space of an m n × matrix A.
For a help with this exercise, please refer to Lectures 28, 30, and 31.
Theory: When the columns of A are linearly independent, a vector p is the orthogonal
projection of a vector b∉ Col A onto the Col A if and only if p x = Aˆ , where xˆ is the unique
least-squares solution of Ax b = , or equivalently, the unique solution of the normal equations
T T AA A x b = .
Also, if p is the orthogonal projection of a vector b onto Col A, then there exists a unique
vector z, such that, z bp = − and z is orthogonal to Col A.
**Your program should allow a possibility that the columns of A are not linearly independent.
In order for the algorithm to work, we will use the function shrink() to create a new matrix,
also denoted A, whose columns form a basis for Col A. Thus, we assign in our code:
A=shrink(A);
The function shrink() was created in Exercise 2 of Project 3 and you should have it in your
Current Folder in MATLAB. If not, please see the code below:
function B=shrink(A)
format compact
[~,pivot]=rref(A);
B=A(:,pivot);
end
You will create a function proj as defined below. The inputs are an m n × matrix A and a
column vector b. The outputs will be the projection p of the vector b onto the Col A and the
vector z which is the component of b orthogonal to the Col A.
**Your function will begin with:
function [p,z]=proj(A,b)
format compact
A=shrink(A);
m=size(A,1);
6
**First, the program checks whether the input vector b has exactly m entries, where m is the
number of rows in A. If it doesn’t, the program breaks with a message ‘No solution:
dimensions of A and b disagree’, outputs and displays the empty vectors p and z.
If b has exactly m entries, we proceed to the next step:
**Determine whether it is the case that b∈ Col A. You should use a MATLAB function
rank(). If b∈ Col A, you will just assign the values to the vectors p and z based on the
general theory (no computations are needed!) and output a message ‘b is in Col A’. After
that, the program terminates.
If b∉ Col A, proceed to the next step:
**Determine whether it is the case that b is orthogonal to Col A. Due to round-off errors in
MATLAB computations, you will need to use closetozeroroundoff with p = 7 to check
orthogonality. If b is orthogonal to Col A, you should just assign the values to the vectors p
and z (no computations are needed!) and output a message ‘b is orthogonal to Col A’.
After that, the program terminates.
**If b is not orthogonal to Col A, proceed to the next steps:
Find the solution of the normal equations (see the Theory above), a vector x (use the inverse
matrix or the backslash operator, \ ). Output the least squares solution, x, with a message:
‘the least squares solution of the system is’
(display x).
**Then, calculate and output a vector
x1=A\b
and check, by using the function closetozeroroundoff with p=12, whether the vectors x and
x1 match within the given precision. If yes, output a message: ‘A\b returns the leastsquares solution of an inconsistent system Ax=b’
**Next, calculate the vectors p and z (see the Theory above).
**Check whether the vector z is, indeed, orthogonal to the Col A – the function
closetozeroroundoff with p=7 should be used here. If your code confirms that, display a
message: ‘z is orthogonal to Col A. Great job!’ Otherwise, output a message like:
‘Oops! Is there a bug in my code?’ and terminate the program.
**Finally, use the vector z to compute the distance d from b to Col A. Output d with the
message:
fprintf(‘the distance from b to Col A is %i’,d)
Hint: use a MATLAB built in function norm().
This is the end of the function proj.
**Print the functions closetozeroroundoff, shrink, proj in your Live Script.
**Run the function [p,z]=proj(A,b)on the following choices of matrices A and vectors b.
Notice that some of the matrices are created in several steps and some intermediate outputs
have been suppressed.
%(a)
A=magic(4), b=sum(A,2)
%(b)
A=magic(4); A=A(:,1:3),b=(1:4)’
7
%(c)
A=magic(6), E=eye(6); b=E(:,6)
%(d)
A=magic(6), b=(1:5)’
%(e)
A=magic(5), b = rand(5,1)
%(f)
A=ones(4); A(:)=1:16, b=[1;0;1;0]
%(g)
B=ones(4); B(:)=1:16, A=null(B,’r’), b=ones(4,1)
%Analyze the outputs in parts (e) and write a comment that would explain a reason why a
random vector b belongs to the Col A.
BONUS! (1 point)
%For part (g), based on the input matrix A and the outputs, state to which vector space the
vector b has to belong.
**You will need to present some additional computation in your Live Script to support your
responses. Supply them with the corresponding output messages and/or comments.
The Exact and Least-Squares Solutions
Please review the section Theory below very carefully – you will be using these facts when
working on Exercise 4.
Theory: An m n × matrix U has orthonormal columns if and only if T UU I = n , where n I is an
n n × identity matrix. If U is a square matrix with orthonormal columns, it is called orthogonal.
We will work with the system Ax=b, where the columns of A are linearly independent.
When solving a consistent system Ax=b (that is, b∈Col A), the unique “exact” solution can be
found by using the backslash operator: x =A\b.
When solving a consistent system Ax=b, where A is a matrix with orthonormal columns, we
can find the unique solution x in two ways as indicated below:
(1) using the backslash operator, \ , that is, x =A\b;
(2) using the Orthogonal Decomposition theorem (see Lecture 28).
When solving an inconsistent system Ax=b (that is, b∉ColA), we will be looking for the
least-squares solution, which can be found either by solving the normal equations,
T T AA A x b = , or by using the backslash operator, A\b (see Exercise 3 of this Project).
Another way of calculating the least squares solution of an inconsistent system Ax=b is to
employ an orthonormal basis U for the Col A. If the columns of an m n × matrix U form an
orthonormal basis for Col A, then, for any vector m b∈ , we can find its projection bˆ onto the
8
Col A by the formula ˆ T b b =UU , and, then, calculate the least-squares solution x
 as the
“exact” solution of a consistent system Ax b = 
.
Using the projection of b onto the Col A, which could be represented either as Ax
 or as bˆ , we
can calculate the least-squares error of approximation of the vector b by the elements of the
Col A as norm A (b x − )  or as norm(b b− )
 (either of them is also the distance from b to Col A
– see Exercise 3 of this Project).
Exercise 4 (5 points) Difficulty: Hard
In this exercise, you will work with the matrix equation Ax b = , where the columns of A are
linearly independent. You will look for the unique “exact” solution, for a consistent equation
(b∈Col A); and you will look for the unique least-squares solution, for an inconsistent
equation (b∉Col A).
**Create a function in MATLAB that begins with
function X=solvemore(A,b)
format compact
format long
[m,n]=size(A);
The inputs are an m n × matrix A, whose columns form a basis for Col A, and a vector m b∈ .
Some of the input matrices A will be created by running the function shrink() – it will leave
only linearly independent columns. You should have already had the function shrink()
created in a file in MATLAB, if not, please see the code in Exercise 3 of this Project.
**Next step is to determine if b∈Col A or b∉Col A – please use a MATLAB function
rank() – and you will employ a logical “if … else” statement to work with the two possible
outcomes as indicated below.
This is the beginning of “if … else” statement:
“If” b is in Col A
first, we output a message: ‘The system is consistent – look for the exact
solution’ and go through the following steps:
**Determine whether the matrix A has orthonormal columns and, if it is the case, determine if
A is also orthogonal.
Hint: To conduct this test in MATLAB, you will need to use the function
closetozeroroundoff() with p = 7.
**If A does not have orthonormal columns, output a corresponding message and find the
unique solution x1 of Ax b = using the backslash operator, \ . In this case, we assign X=x1 and
terminate the program.
**If A does have orthonormal columns, output a corresponding message, and your code will
continue with the following tasks:
First, it will check whether A is orthogonal. If yes, it outputs a message ‘A is orthogonal’.
9
If not, the output message should be ‘A has orthonormal columns but is not
orthogonal’.
Next, output the solution of Ax b = in two ways (do not display the outputs x1 and x2 here):
(1) using the backslash operator, \ , denote this solution x1,
(2) using the Orthogonal Decomposition theorem (Lecture 28), denote this solution x2.
Hint: for part (2), you can either use a “for loop” to calculate the entries of the vector x2, one
by one, or you can employ the transpose of A and output x2 by using a MATLAB operation.
else (b is not in Col A)
we display a message ‘The system is inconsistent: look for the least-squares
solution’ and, then, we will find the least-squares solution in both ways as indicated below:
**First way: we will find the unique least-squares solution directly by either solving the
normal equations or using the backslash operator. We denote the solution x1 and display it
with a message that it is the least-squares solution of the system.
**Second way: we will use an orthonormal basis for Col A to find the least-squares solution. It
might be possible that the matrix A has orthonormal columns and, if not, we will create an
orthonormal basis for Col A.
Check if the matrix A has orthonormal columns. If yes, your outputs will be:
disp(‘A has orthonormal columns: orthonormal basis for Col A is U=A’)
U=A
If the matrix A does not have orthonormal columns, find an orthonormal basis U for Col A
using a MATLAB function orth(). Output and display U with a message
disp(‘A does not have orthonormal columns: orthonormal basis for Col A is’)
U=orth(A)
Next, using the orthonormal basis U for Col A, calculate the projection of b onto Col A, denote
it b1. Output and display vector b1 with the corresponding message.
Next, find the least-squares solution x2 as the “exact” solution of the equation Ax = b1 (you
can use the backslash operator here). Output x2 with the message: ‘The least-squares
solution using the projection b1 is’ (output and display x2).
**Use the least-squares solution x1 to find a least-squares error of approximation of b by the
elements of Col A, denote it n1.
Display n1 with a message:
fprintf(‘Error of approximation of b by vector A*x1 of Col A is\n’)
n1
**Now, we use the projection b1 to find the least-squares error of approximation of b by the
elements of Col A, denote it n2.
Display n2 with a message:
fprintf(‘Error of approximation of b by vector b1 of Col A is\n’)
n2
**Next, input a vector x=rand(n,1). Compute an error n3=norm(b-A*x) of approximation
of the vector b by a vector Ax of the Col A for a random vector x.
Output it with a message:
fprintf(‘error of approximation of b by A*x of Col A for random x is\n’)
n3
This is the end of your “if … else” statement
10
After completing all of the above
your code will proceed with composing the output X for the cases when we have two solutions
x1 and x2 for both consistent and inconsistent systems, but, first, we will check if the
corresponding entries of the two solutions x1 and x2 are sufficiently close to each other. We
will run the function closetozeroroundoff() with p = 12 on the vector of the difference
between x1 and x2, and compare the output with the zero vector. If your code confirms that
the corresponding entries of the solutions are in the range of 10^(-12) from each other, output a
message ‘solutions x1 and x2 are sufficiently close to each other’, and assign
(and display) X=[x1,x2]. Otherwise, the output message should be ‘Check the code!’ and
the empty matrix should be assigned to the output X.
**Print the functions closetozeroroundoff, shrink, solvemore in your Live Script.
**Run the function X=solvemore(A,b) on the following matrices. Please type (or copy and
paste) the matrices and the vectors exactly as they appear below.
%(a)
A=magic(4); b=A(:,4), A=orth(A)
%(b)
A=magic(5); A=orth(A), b=rand(5,1)
%(c)
A=magic(6); A=shrink(A), b=ones(6,1)
%(d)
A=magic(6); A=shrink(A), b=rand(6,1)
%(e)
A=magic(4); A=orth(A), b=rand(4,1)
% Compare the outputs n1 and n2 and write a comment whether they are close to each other.
% Next, compare n1 with n3. Write a comment whether the least-squares solution, x1 (or x2)
may, indeed, minimize the distance between a vector b and the vectors Ax of the Col A (which
has to be true according to the general theory).
Part III. Application to Polynomials
Applications to Linear Models – Regression Lines
In this part of the project, you will write codes that output the least-squares fit equation for the
given data. The equation is determined by the parameter vector c (that has to be calculated),
the design matrix X, and the observation vector y, such that, the 2-norm of the residual
vector e = y – Xc is the minimum.
Assume that we are given the data points ( x y i i , ), where i m =1: . A choice of the equation of a
curve which gives the least-squares fit to the data is, in general, arbitrary. First, we will take a
look at the equation of a straight line 1 0 y cx c = + that minimizes the sum of squares of the
residuals – it is called the least-squares regression line. The parameter vector c can be
determined by computing the least-squares solution of Xc y = , where
11
1
2
1
1
1 m
x
x
X
x
 
 
  =  
   
  , 1
0
c
c
  =    
c ,
1
2
m
y
y
y
 
 
  =  
   
y  .
Exercise 5 (3 points) Difficulty: Easy
**First, create the following function in MATLAB that will be used in the function below:
function []=polyplot(a,b,p)
x=(a:(b-a)/50:b)’;
y=polyval(p,x);
plot(x,y);
end
**Create another function in MATLAB that begins with the following lines:
function c=lstsqline(x,y)
hold off
format
format compact
x=x’;
y=y’;
a=x(1);
m=length(x);
b=x(m);
disp(‘the design matrix is’)
X=[x,ones(m,1)]
disp(‘the parameter vector is’)
c=lscov(X,y)
disp(‘the norm of the residual vector is’)
N=norm(y-X*c)
plot(x,y,’*’),hold on
polyplot(a,b,c’);
fprintf(‘the least-squares regression line is\n’)
P=poly2sym(c)
Note: The data vectors x and y (inputs) are the row vectors – we use the transpose function in
the code above to convert them into the column vectors. The parameter vector p in the function
polyplot(a,b,p) has to be a row vector – in the code above it appears as c’.
**The MATLAB command c=lscov(X,y)calculates the parameter vector c which is, in fact,
the least-squares solution of the inconsistent system Xc y = . You will continue the function
lstsqline with a verification of this fact: output (do not display) the least-squares solution c1
of the system Xc y = , calculated by any of the method from Exercise 4 of this Project, and,
then, employ closetozeroroundoff() with p=7, to verify that c and c1 match. If it is the
case, output the message ‘c is the least-squares solution’
(Please make sure that you will receive this message after running the function)
**Complete your function lstsqline with the commands:
hold off
end
12
**Print the functions polyplot and lstsqline in your Live Script.
**Input the vectors:
x = [0,2,3,5,6], y = [1,4,3,4,5]
**Run the function
c=lstsqline(x,y);
Your outputs have to be: the design matrix X, the parameter vector c, the 2-norm of the residual
vector N, the equation of the least-squares regression line P, the plot that contains both the data
points and the line of the best fit, and the message confirming that c is, indeed, the leastsquares solution.
Exercise 6 (4 points) Difficulty: Moderate
In this Exercise you will find a polynomial of degree n of the best fit for the given data points.
**Create a new function in a file
function c=lstsqpoly(x,y,n)
which takes as inputs the data vectors x and x and a positive integer number n.
The function lstsqpoly(x,y,n) is a modification of the function lstsqline(x,y) from
Exercise 5 in the way that it has to output the least-squares polynomial of degree n.
You will take the function c=lstsqline(x,y)and make the three changes in it as indicated
below (everything else will stay the same):
**Add one more input variable n and re-name the function as lstsqpoly.
**Replace the matrix X with the new matrix, also named X, whose form depends on the degree
n of the polynomial that we use to fit the data. For example, when n=3, the least-squares
polynomial will be of degree 3, and the design matrix X will have a form:

3 2
111
3 2
222
3 2
1
1
1 mmm
xxx
xxx
X
xxx
 
 
  =  
 
   
   .
**Replace the output message for the polynomial P with a message
fprintf(‘the polynomial of degree %i of the best least-squares fit is\n’,n)
Please make sure that you have the function polyplot in your Current Folder in MATLAB – it
was used in the function lstsqline and it will be used in the function lstsqpoly as well.
**Print the functions polyplot and lstsqpoly in the Live Script.
**Run the function c=lstsqpoly(x,y,n) on the vectors from Exercise 5:
x = [0,2,3,5,6], y = [1,4,3,4,5]
for each n = 1, 2, 3, 4.
For each n, the outputs have to be: the design matrix X, the parameter vectors c, the 2-norm of
the residual vector N, the equation of the least-squares polynomial P, the plot containing both
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the data points and the graph of the polynomial P, and the message confirming that c is,
indeed, the least-squares solution.
.
% Verify that your code in Exercise 6 for n = 1 is consistent with the code in Exercise 5, that
is, the outputs of the functions c=lstsqline(x,y) and c=lstsqpoly(x,y,1) match. Write a
BONUS! (1 point)
% Analyze the outputs for n = 4. Justify the fact that the least-squares polynomial of degree 4
is, actually interpolates the 5 data points.
Part III. Application to Dynamical Systems
In this part of the Project, you will be working with application of eigenvalues and
eigenvectors to the description of evolution of a discrete dynamical system.
Exercise 7 (5 points) Difficulty: Very Hard
Theory: We will work with a 2×2 matrix A, whose entries are real numbers, and the related
linear difference equation
1 ( 0,1, 2,…) k k A k x x + = = (1)
that describes evolution of a discrete dynamical system with an initial vector 0 x in 2  .
If A is diagonalizable, there exists an eigenvector basis 1 2 {, } v v for 2  , and, for any initial
vector 0 x in 2  , there is a unique vector of weights C cc = ( 1 2. , ), such that,
0 11 2 2 xvv = + c c . (2)
Moreover, the solution of the difference equation (1) is represented in an explicit form as
11 1 2 2 2 ( ) ( ) ( 0,1, 2,…) k k
k xvv =+ = cc k λ λ (3)
where λ1 and λ2 are the eigenvalues of A corresponding to the eignevectors 1 v and 2 v ,
respectively.
To represent the solution geometrically:
we will use equation (3) to calculate 123 xxx , , ,…, when working with the matrices whose
eigenvalues are positive numbers;
and we will use the recurrence relation (1) to find the consecutive 123 xxx , , ,…, when working
with the matrices whose eigenvalues are complex (non-real) numbers.
The graph of 012 xxx , , ,… is called the trajectory of the dynamical system.
We will accept the following description of the origin in relation to the patterns of the
trajectories of a dynamical system:
I. All trajectories tend towards the origin. In this case, the origin is called an attractor, and
it happens when both eigenvalues are less than 1 in magnitude. The direction of the
greatest attraction is through the origin and the eigenvector corresponding to the
eigenvalue of the smaller magnitude.
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II. All solutions, except for the constant (zero) solution, are unbounded, and their
trajectories tend away from the origin. In this case, the origin is called a repeller, and it
happens when both eigenvalues are greater than 1 in magnitude. The direction of the
greatest repulsion is through the origin and the eigenvector corresponding to the
eigenvalue of the larger magnitude.
III. Some trajectories tend towards the origin and the others tend away from the origin. In
this case, the origin is called a saddle point, and it happens when one of the eigenvalues
is less than 1 in magnitude and the other one is greater than 1 in magnitude. The
direction of the (greatest) attraction is through the origin and the eigenvector
corresponding to the eigenvalue of the smaller magnitude; and the direction of the
greatest repulsion is through the origin and the eigenvector corresponding to the
eigenvalue of the larger magnitude.
For more details, please refer to Lecture 26 and Section 5.6 of the Textbook.
**Create a function in MATLAB that begins with the lines:
function []=trajectory(A,X0,N)
format
format compact
L=eig(A);
The inputs are: a 2×2 real-valued matrix A, a matrix X0 whose columns are the initial vectors in
2  , and the number N of the vectors 123 , , ,…, N xxx x which we will calculate.
We will use a logical ‘if … else” statement to break up the whole code into two parts:
Part 1
“if” matrix A has real eigenvalues
first, we display a message ‘the eigenvalues of A are real’ and proceed with the
**Check if A has a zero eigenvalue. If it is the case, output a message ‘A has a zero
eigenvalue’ and terminate the program – we will not graph the solutions in this case.
**You will continue your code for trajectory by modifying the function eigen, which was
created in Exercise 1 of this Project. The details how the function eigen should be modified to
be included into the function trajectory are listed below – please remove/suppress all the
tasks/outputs of the function eigen that are not on the list.
Important: Please notice that you have n=2 in your code and you also need to replace the
command null( ), used in the function eigen, with null( ,’r’) to output a “rational” basis
for an eigenspace.
After adjusting the function eigen for your code, you will add some more commands that are
listed below in these instructions.
The function trajectory has to contain the following parts coming from the function eigen:
**Output a row vector L of the sorted eigenvalues of A, where the multiple eigenvalues are all
equal to each other: the entries of L have to be written in the ascending order and each
eigenvalue repeats as many times as its multiplicity. Display L with a message:
fprintf(‘all sorted eigenvalues of A are\n’)
L
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**Output (do not display) the matrix P whose columns are bases for the eigenspaces
corresponding to the sorted eigenvalues in L.
**Check if A is diagonalizable.
If A is not diagonalizable, output a message: ‘A is not diagonalizable: there is no
eigenvector basis for R^2’
and terminate the code.
If A is diagonalizable, output the following:
fprintf(‘A is diagonalizable: there exists an eigenvector basis for R^2\n’)
fprintf(‘it is formed by V1,V2 corresponding to sorted eigenvalues in L\n’)
(assign to V1 and V2 the columns 1 and 2 of P, respectively, and display V1 and V2 here.)
This is the end of the part of your code which relates to the function eigen.
Continue your function trajectory with the following:
**Check if all eigenvalues of A are positive numbers. If at least one of the eigenvalues of A is
negative, we terminate the code with a message ‘A has a negative eigenvalue’
**Next (for the matrices whose eigenvalues are positive numbers), we will determine whether
the origin is an attractor, a repeller, or a saddle point. (See the Theory above.) Program an
output message for each of the three cases and also output a message that will state the
direction of the greatest attraction/repulsion.
An example of the output messages when the origin as an attractor, is below:
disp(‘the origin is an attractor’)
fprintf(‘a direction of greatest attraction is through 0 and\n’)
V1
(Remember, V1 and V2 are the vectors in an eigenvector basis for 2  corresponding to the
positive eigenvalues of A sorted in the ascending order.)
**Then, we will check if at least one of the eigenvalues of A is the integer number 1. If it is the
case, we display a message ‘A has an eigenvalue 1′ and go to the next step (do not
terminate the code here).
Next, we will create plots of the trajectories of a dynamical system.
**First, you will introduce a new matrix of the initial vectors, which is a horizontal
concatenation of the vectors V1,V2,-V1,-V2 with the input matrix X0.Type in your code:
X0 = [V1,V2,-V1,-V2,X0];
**Next, type the following lines in your code:
n=size(X0,2);
X=zeros(2,N+1);
The first line counts the total number of the initial vectors, and the second line stores a matrix
X, whose entries will be re-calculated by using a double “for loop” as indicated below:
**For each initial vector X0(:,i), where i=1:n, calculate the coefficient vector C according to
the formula (2). Use the vector C to calculate by formula (3) the N+1 vectors 012 , , ,…, N xxx x
and output them as N+1 columns of the matrix X. Proceed with graphing the trajectories.
Type (see also Exercise 5 of Project 2):
x=X(1,:);y=X(2,:);
plot(x,y,’*’), hold on
plot(x,y)
16
and program the viewing window as
v=[-1 1 -1 1];
for the two cases: when the origin is an attractor and when A has an eigenvalue 1;
and program the viewing window as
v=[-5 5 -5 5];
for all other cases.
After that, type:
axis(v)
This is the end of the double “for loop”.
This is the end of the “IF” part of your code – proceed to the “ELSE” part.
Part 2
else (matrix A has complex conjugate (non-real) eigenvalues)
first, we output a message
disp(‘the eigenvalues of A are complex conjugate (non-real) numbers’)
**Next, we will graph the trajectories by calculating consecutive iterations 123 , , ,…, N xxx x , for
each initial vector 0 x (a column of the input X0) by using the recurrence equation (1) (see the
Theory).
Begin this part with the commands:
L=eig(A)
magn=abs(L)
n=size(X0,2);
X=zeros(2,N+1);
Entries of the vector magn are the magnitudes (moduli) of the complex eigenvalues in L.
**Then, use a double “for loop” to graph the trajectory:
for each initial vector X0(:,i), where i=1:n, calculate the consecutive iterations
012 , , ,…, N xxx x by formula (1), and output them as the N+1 columns of X.
Plot the trajectory by the following lines:
x=X(1,:);y=X(2,:);
plot(x,y,’*’), hold on
plot(x,y)
This is the end of a double “for loop”.
This is the end of the “ELSE” part of your “if … else” statement.
**The last commands in your function trajectory are:
hold off
end
**Print the function trajectory in your Live Script.
**Run the function trajectory(A,X0,N) on the following sets of variables.
%(a)
A=[2 0;0 .5]
X0=[[.1;5],[-.1;5],[-.1;-5],[.1;-5]];
N=10;
17
%(b)
A=[2 0; 0 3]
X0=[[0;0],[1;1],[1;-1],[-1;-1],[-1;1],[1;.1],[-1;.1],[-1;-.1],[1;-.1]];
N=10;
A=[.80 0;0 .64]
X0=[[1;1],[-1;1],[-1;-1],[1;-1],[.5;1],[-.5;1],[-.5;-1],[.5;-1]];
N=10;
%(c)
A=[.80 0;0 .64]
X0=[[1;1],[-1;1],[-1;-1],[1;-1],[.5;1],[-.5;1],[-.5;-1],[.5;-1]];
N=10;
%(d)
A=[.64 0;0 .64]
X0=[[1;1],[-1;1],[-1;-1],[1;-1],[.5;1],[-.5;1],[-.5;-1],[.5;-1]];
N=10;
%(e)
A=[5 0; 1 5]
X0=[[1;1],[-1;1],[-1;-1],[1;-1]];
N=10;
%(f)
A=[1 0;0 .64]
X0=[[.5;1],[-.5;1],[-.5;-1],[.5;-1]];
N=10;
%(g)
A=[.90 .04;.10 .96]
X0=[[.2;.8],[.1;.9],[.9;.1],[.6;.4],[.5;.5]];
N=10;
%(h)
A=[0 -1;1 0]
X0=[[.1;.1],[.5;.5],[.8;0],[1;-1]];
N=10;
%(i)
A=[.5 -.6;.75 1.1]
X0=[[.1;.1],[.5;.5],[-1;-1],[1;-1]];
N=10;
%(j)
A=[.8 .5; -.1 1.0]
X0=[[1;0],[0;-1]];
N=100;
%(k)
A=[1.01 -1.02;1.02 1.01]
X0=[[1;0],[0;-1]];
N=10;
%(l)
A=[.3 .4;-.3 1.1]
X0=[[0;.5],[1;1],[-1;-1],[0;-.5],[-1;-.8],[1;.8],[-.5;.5],[.5;-.5]];
N=10;
%(m)
A=[.5 .6; -.3 1.4]
X0=[[.1;2],[4;0],[-4;0],[-.1;-2]];
N=30;
18
%(n)
A=[.8 .3; -.4 1.5]
X0=[[0;1],[-1;0],[0;-1],[1;0],[0;0],[3;0],[-3;0]];
N=50;
%(p)
A=[[1 2;2 4]]
X0=[];
N=10;
%(q)
A=[-.64 0;0 1.36]
X0=[];
N=10;
BONUS! (1 point)
%Analyze the outputs for the choice (d) and use equation (3) to explain why all trajectories are
straight lines.
%Explain the pattern of behavior of the solutions for part (f).
BONUS! (1 point)
In part (g), we have a regular stochastic matrix A. Slightly modify your function trajectory
and the vector X0 to output additionally the unique steady-state vector q that describes the
evolution of the given Markov Chain. Save your new function in MATLAB as trajectory_1
**Print the function trajectory_1
**Run trajectory_1(A,X0,N) with N=100 on the stochastic matrix A given in (g) and your
own input X0. Output and display the steady-state vector q.
BONUS! (1 point)
%Analyze the outputs for choices (h),(i),(j), and (k), where the eigenvalues are complex
conjugates. Use the equation (3) and the output for magn to explain the pattern of the
trajectories for each system.
You are done with the Projects!!!

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