(1) What is the rank of the matrix A?

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

The rank of a Matrix is the maximum number of independent rows (or, the maximum number of independent columns). A square matrix \(A_{n×n}\) is non-singular only if its rank is equal to n.

The process for computing the rank involves translating a matrix into echelon form. Putting this matrix into echelon form will be a tedious process but we will go through each and every step.

Add the first row to the second row \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} \]

Add -5 times row 1 to row 4 \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix} \] Divide row 2 by 2 \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix} \]

subtract the second row from the first row \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & -6 & -17 & -23 \end{bmatrix} \]

6 times row 2 plus row 4 \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & 0 & -5 & -2 \end{bmatrix} \]

Divide row 3 by -4 \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -5 & -2 \end{bmatrix} \]

5 times row 3 plus row 4 \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 9/8 \end{bmatrix} \]

divide row 4 by 9/8 \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Add -5/8 row 4 to row 3 \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

-7/2 row 4 plus row 2 \[ A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] add-4 times row 1 to row 4 \[ A=\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

-2 times row 3 plus row 2 \[ A=\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] We can bundle the next row operations since rows 2-4 will remain unchanged -3 times row 3 to row 1 -2 times row 2 plus row 1 \[ A=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

The rank of matrix A is 4 since there are 4 non zero linearly independent rows

#Let's store information in Matrix
A <- matrix( 
   c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), # the data elements 
   nrow=4,              # number of rows 
   ncol=4,              # number of columns 
   byrow = TRUE)        # fill matrix by rows 

qr(A)$rank

[1] 4

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

We know that let \(rank(A_{m*n})\) = rank of matrix A, where matrix A is an m x n matrix and m>n \[ rank(A_{m*n})\le \min(m,n) \] n if all n columns are linearly independent. \[ rank(A_{m*n})\le n \] Minimum Rank for m *n where m >n is 1 as matrix is non-zero ; it will have one linear independent columns. \[ rank(A_{m*n})\ge 1 \] Thus the minimum rank of matrix A would be 1

(3) What is the rank of matrix B?

\[ B=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix} \] We need to follows the same procedure as problem 1 to find the rank of matrix B.

-3 times row 1 plus row 2 \[ B=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 2 & 4 & 2 \end{bmatrix} \]

-2 times row 1 plus row 3 \[ B=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] Thus, The rank of matrix B is 1

#In R
#Let's store information in Matrix
B <- matrix( 
   c(1,2,1,3,6,3,2,4,2), # the data elements 
   nrow=3,              # number of rows 
   ncol=3,              # number of columns 
   byrow = TRUE)        # fill matrix by rows 

qr(B)$rank

[1] 1

Both Answers are same

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.

\[ A=\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \] As we know \(Ax=\lambda x\\det(A-I\lambda)=0\),

\[ \begin{aligned} \begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{vmatrix} - \lambda \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} &= \begin{vmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{vmatrix} \\ &=(1-\lambda)\begin{vmatrix} 4-\lambda & 5\\ 0 & 6-\lambda \end{vmatrix} -2 \begin{vmatrix} 0 & 5 \\ 0 & 6-\lambda \end{vmatrix} + 3 \begin{vmatrix} 0 & 4-\lambda \\ 0 & 0 \end{vmatrix} \\ &=(1- \lambda)((4- \lambda)(6-\lambda))+2(0-0)+3(0-0)\\ &=(1- \lambda)(4-\lambda)(6-\lambda)-0+0\\ &=(1-\lambda)(24-4\lambda-6\lambda+\lambda^{2})\\ &=(1-\lambda)(24-10\lambda+\lambda^{2})\\ &=24-10\lambda+\lambda^{2}-24\lambda+10\lambda^{2}-\lambda^{3}\\ &=-\lambda^{3}+11\lambda^{2}-34\lambda+24\\ &=\lambda^{3}-11\lambda^{2}+34\lambda-24 \end{aligned} \] Characteristic polynomial of the matrix is

\(pA(x)=x^3-11x^2+34x-24\)

Let’s Verified

#Let's store information in Matrix
A <- matrix( 
   c(1,2,3,0,4,5,0,0,6), # the data elements 
   nrow=3,              # number of rows 
   ncol=3,              # number of columns 
   byrow = TRUE)        # fill matrix by rows 

#polynomial of the matrix
charpoly(A)

[1]   1 -11  34 -24

Thus, characteristic equation is verified

Lets use R to verify our eigenvalues

#
e <- eigen(A)

#Eigen Values
e$values

[1] 6 4 1

#Eigen Vectors
e$vectors

          [,1]      [,2] [,3]
[1,] 0.5108407 0.5547002    1
[2,] 0.7981886 0.8320503    0
[3,] 0.3192754 0.0000000    0