Assignment #3

Problem Set 1

Question 1

What is the rank of the matrix A?

\[A\quad =\quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix}\]

Answer: To find the rank of a matrix, we can either find the maximum number of linearly independent column vectors in the matrix or the maximum number of linearly independent row vectors in the matrix.

The maximum number of linearly independent vectors in a matrix is equal to the number of non-zero rows in its row echelon matrix. Therefore, to find the rank of a matrix, we simply transform the matrix A to its row echelon form and count the number of non-zero rows.

The rank of the matrix A is 4. This is found by the two ways demonstrated as below.

1. Using a rank of matrix calculation package, “matrixcalc”.

library(matrixcalc)

mtx_A <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), byrow=T, nrow=4, ncol=4)
matrix.rank(mtx_A)

## [1] 4

2. Using the row echelon form to find the number of linearly independent row. We can see that the number of non-zero rows in the row echelon form below is 4.

Row Operation 1: add 1 times the 1st row to the 2nd row

\[\begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix}\quad\]

Row Operation 2: add -5 times the 1st row to the 4th row

\[\quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix}\quad\]

Row Operation 3: multiply the 2nd row by 1/2

\[\quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix}\quad\]

Row Operation 4: add -1 times the 2nd row to the 3rd row

\[\quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & -6 & -17 & -23 \end{bmatrix}\quad\]

Row Operation 5: add 6 times the 2nd row to the 4th row

\[\quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & -6 & -17 & -23 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & 0 & -5 & -2 \end{bmatrix}\quad\]

Row Operation 6: multiply the 3rd row by -1/4

\[\quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & 0 & -5 & -2 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -5 & -2 \end{bmatrix}\quad\]

Row Operation 7: add 5 times the 3rd row to the 4th row

\[\quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -5 & -2 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 9/8 \end{bmatrix}\quad\]

Row Operation 8: multiply the 4th row by 8/9

\[\quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 9/8 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 1 \end{bmatrix}\quad\]

Question 2

Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

Answer:

For m x n matrix, if m is greater than n, then the maximum rank of the matrix is n. For the case where m x n matrix, the maximum rank is always min(m x n).

The maximum number of linearly independent rows in a matrix is called the row rank, and the maximum number of linearly independent columns in a matrix is called the column rank. Because for any matrix, the row rank = the column rank; therefore, for the case of a square matrix, the maximum rank is less than or equal to the number of row or column of that matrix.

Even there is only one element in a matrix, its minimum rank would be one. The rank of a matrix would be zero only if the matrix had no elements.

Question 3

What is the rank of matrix B?

\[B\quad =\quad \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix}\]

Answer: To find the rank of a matrix, we can either find the maximum number of linearly independent column vectors in the matrix or the maximum number of linearly independent row vectors in the matrix.

The maximum number of linearly independent vectors in a matrix is equal to the number of non-zero rows in its row echelon matrix. Therefore, to find the rank of a matrix, we simply transform the matrix A to its row echelon form and count the number of non-zero rows.

The rank of the matrix A is 1. This is found by the two ways demonstrated as below.

1. Using a rank of matrix calculation package, “matrixcalc”.

library(matrixcalc)

mtx_B <- matrix(c(1,2,1,3,6,3,2,4,2), byrow=T, nrow=3, ncol=3)
matrix.rank(mtx_B)

## [1] 1

2. Using the row echelon form to find the number of linearly independent row. We can see that the number of non-zero rows in the row echelon form below is 1.

Row Operation 1: add -3 times the 1st row to the 2nd row

\[\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 2 & 4 & 2 \end{bmatrix}\quad\]

Row Operation 2: add -2 times the 1st row to the 3rd row

\[\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 2 & 4 & 2 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\quad\]

Problem Set 2

Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work. You’ll need to write out the characteristic polynomial and show your solution.

\[A\quad =\quad \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}\]

Answer:

Let lambda, \(\lambda\) is an eigenvalue of A if and only if \(A\vec { \nu } =\quad \lambda \vec { \nu }\) for some non-zero

\(A\vec { \nu } = \lambda \vec { \nu }\) is true if and only if \(0 = \lambda \vec { \nu } - A\vec { \nu } = (\lambda { I }_{ n } - A)\vec { \nu }\) . \((\lambda { I }_{ n } - A)\vec { \nu } = 0\) is true if and only of \(det(\lambda { I }_{ n } - A) = 0\) for some non-zero \(\vec { \nu }\).

Applying the \(det(\lambda { I }_{ n } - A) = 0\) to find the eigenvalue of matrix A.

\[\lambda { I }_{ 3 }=\begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}\quad \longrightarrow \quad \lambda { I }_{ 3 }-A=\begin{bmatrix} \lambda -1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda -6 \end{bmatrix}\]

Using the Sarrus’ rule to find the \(det(\lambda { I }_{ 3 } - A)\).

\[\begin{bmatrix} \lambda -1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda -6 \end{bmatrix}\begin{matrix} \lambda -1 & -2 \\ 0 & \lambda -4 \\ 0 & 0 \end{matrix}\]

\(det(\lambda { I }_{ 3 } - A) = (\lambda-1)(\lambda-4)(\lambda-6) + (-2)(-5)(0) + (-3)(0)(0) - (0)(\lambda-4)(-3) - (0)(-5)(\lambda-1) - (\lambda-6)(0)(-2)\)

= \((\lambda-1)(\lambda-4)(\lambda-6)\)

So, the characteristic of polynomial for the matrix A is

\[pA(\lambda)=(\lambda-1)(\lambda-4)(\lambda-6)\]

The determinant of this matrix for any lambda has to be equal to zero if any only if lambda is truly an eigenvalue. So, let set \((\lambda-1)(\lambda-4)(\lambda-6)=0\). Therefore, the possible eigenvalues of matrix A are:

\[ {\displaystyle \begin{aligned} \lambda=1,\quad \lambda=4,\quad \lambda=6 \end{aligned} } \]

Now, let find the corresponding eigenvectors of A using the equation \((\lambda { I }_{ n } - A)\vec { \nu }=0\)

\(~\)

1. Let take the case of \(\lambda=1\),

\[\begin{bmatrix} \lambda -1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda -6 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1-1 & -2 & -3 \\ 0 & 1-4 & -5 \\ 0 & 0 & 1-6 \end{bmatrix}\vec { \nu } =0\quad \longrightarrow \quad \begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix}\vec { \nu } =0\]

The null space of this matrix \(\begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix}\) is the eigenspace. To solve for eigenvectors, we can do the reduced row echelon form for this matrix.

\(~\)

Row Operation 1: multiply the 1st row by -1/2

\[\begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 0 & 1 & 3/2 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix}\]

Row Operation 2: add 3 times the 1st row to the 2nd row

\[\begin{bmatrix} 0 & 1 & 3/2 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 0 & 1 & 3/2 \\ 0 & 0 & -1/2 \\ 0 & 0 & -5 \end{bmatrix}\]

Row Operation 3: multiply the 2nd row by -2

\[\begin{bmatrix} 0 & 1 & 3/2 \\ 0 & 0 & -1/2 \\ 0 & 0 & -5 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 0 & 1 & 3/2 \\ 0 & 0 & 1 \\ 0 & 0 & -5 \end{bmatrix}\]

Row Operation 4: add 5 times the 2nd row to the 3rd row

\[\begin{bmatrix} 0 & 1 & 3/2 \\ 0 & 0 & 1 \\ 0 & 0 & -5 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 0 & 1 & 3/2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\]

Row Operation 5: add -3/2 times the 2nd row to the 1st row

\[\begin{bmatrix} 0 & 1 & 3/2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\]

\(~\)

\[\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} { v }_{ 1 } \\ { v }_{ 2 } \\ { v }_{ 3 } \end{bmatrix}\quad =\quad \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\quad \longrightarrow \quad { v }_{ 2 }=0,\quad { v }_{ 3 }=0,\quad { v }_{ 1 }={ v }_{ 1 }\]

So, the eigenvector of matrix A associated with the eigenvalue \(\lambda=1\) is:

\[V=\begin{pmatrix} { v }_{ 1 } \\ 0 \\ 0 \end{pmatrix}={ v }_{ 1 }\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\quad for\quad any\quad real\quad number\quad { v }_{ 1 }\neq 0.\] \(~\)

2. Let take the case of \(\lambda=4\),

\[\begin{bmatrix} \lambda -1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda -6 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 4-1 & -2 & -3 \\ 0 & 4-4 & -5 \\ 0 & 0 & 4-6 \end{bmatrix}\vec { \nu } =0\quad \longrightarrow \quad \begin{bmatrix} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{bmatrix}\vec { \nu } =0\]

The null space of this matrix \(\begin{bmatrix} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{bmatrix}\) is the eigenspace. To solve for eigenvectors, we can do the reduced row echelon form for this matrix.

\(~\)

Row Operation 1: multiply the 1st row by 1/3

\[\begin{bmatrix} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & -2/3 & -1 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{bmatrix}\]

Row Operation 2: multiply the 2nd row by -1/5

\[\begin{bmatrix} 1 & -2/3 & -1 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & -2/3 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & -2 \end{bmatrix}\]

Row Operation 3: add 2 times the 2nd row to the 3rd row

\[\begin{bmatrix} 1 & -2/3 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & -2 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & -2/3 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\]

Row Operation 4: add 1 times the 2nd row to the 1st row

\[\begin{bmatrix} 1 & -2/3 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\]

\(~\)

\[\begin{bmatrix} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} { v }_{ 1 } \\ { v }_{ 2 } \\ { v }_{ 3 } \end{bmatrix}\quad =\quad \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\quad \longrightarrow \quad { v }_{ 1 }=2/3{ v }_{ 2 },\quad { v }_{ 3 }=0,\quad { v }_{ 2 }={ v }_{ 2 }\]

So, the eigenvector of matrix A associated with the eigenvalue \(\lambda=4\) is:

\[V=\begin{pmatrix} \frac { 2 }{ 3 } { v }_{ 2 } \\ { v }_{ 2 } \\ 0 \end{pmatrix}={ v }_{ 2 }\begin{pmatrix} \frac { 2 }{ 3 } \\ 0 \\ 0 \end{pmatrix},\quad for\quad any\quad real\quad number\quad { v }_{ 2 }\neq 0.\] \(~\)

3. Let take the case of \(\lambda=6\),

\[\begin{bmatrix} \lambda -1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda -6 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 6-1 & -2 & -3 \\ 0 & 6-4 & -5 \\ 0 & 0 & 6-6 \end{bmatrix}\vec { \nu } =0\quad \longrightarrow \quad \begin{bmatrix} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix}\vec { \nu } =0\]

The null space of this matrix \(\begin{bmatrix} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix}\) is the eigenspace. To solve for eigenvectors, we can do the reduced row echelon form for this matrix.

\(~\)

Row Operation 1: multiply the 1st row by 1/5

\[\begin{bmatrix} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 5 & -2/5 & -3/5 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix}\]

Row Operation 2: multiply the 2nd row by 1/2

\[\begin{bmatrix} 5 & -2/5 & -3/5 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 5 & -2/5 & -3/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\]

Row Operation 3: add 2/5 times the 2nd row to the 1st row

\[\begin{bmatrix} 5 & -2/5 & -3/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\quad \longrightarrow \quad \begin{bmatrix} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\]

\(~\)

\[\begin{bmatrix} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} { v }_{ 1 } \\ { v }_{ 2 } \\ { v }_{ 3 } \end{bmatrix}\quad =\quad \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\quad \longrightarrow \quad { v }_{ 1 }=8/5{ v }_{ 3 },\quad { v }_{ 2 }=5/2{ v }_{ 3 },\quad { v }_{ 3 }={ v }_{ 3 }\]

So, the eigenvector of matrix A associated with the eigenvalue \(\lambda=6\) is:

\[V=\begin{pmatrix} \frac { 8 }{ 5 } { v }_{ 3 } \\ \frac { 5 }{ 2 } { v }_{ 3 } \\ 0 \end{pmatrix}={ v }_{ 3 }\begin{pmatrix} \frac { 8 }{ 5 } \\ \frac { 5 }{ 2 } \\ 0 \end{pmatrix},\quad for\quad any\quad real\quad number\quad { v }_{ 3 }\neq 0.\]