A <- matrix(c(1, 2, 3, 4, -1, 0, 1, 3,0, 1, -2, 1, 5, 4, -2, -3), nrow = 4, byrow = T)
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3qr(A)$rank## [1] 4The Rank of the matrix is 4.
Slol
In a mxn matris where m > n the maximum rank will be the the minimum between m and n. In our case the minumum is n, the rows below the nth column will be multiple of each other and will be eliminated, so the maximum rank will be n.
If we assume the matrix is non-zero the minimum rank will be 1 as there will be atleast 1 pivot.
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix} \]
B <- matrix(c(1, 2, 1, 3, 6, 3, 2, 4, 2), nrow = 3, byrow = T)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2qr(B)$rank## [1] 1The rank of the matrix is 1.
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.
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \] Sol
Eigenvalue
\[ \lambda I_{3} = \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} \]
\[ \lambda I_{3} - A = \begin{bmatrix} \lambda - 1 & -2 & -3 \\ 0 & \lambda - 4 & -5 \\ 0 & 0 & \lambda - 6 \end{bmatrix} \] \(\lambda\) is a eigenvalue of a if, \(det(\lambda I_{3} - A) = 0\)
\((\lambda - 1) (\lambda - 4) (\lambda - 6) + (-2 \times -5 \times 0) + (3 \times 0 \times 0) - (-2 \times 0 \times (\lambda - 6)) - ((\lambda - 1) \times - 5 \times - 0) - (-3 \times (\lambda - 4) \times 0) = 0\)
\(\Rightarrow (\lambda - 1) (\lambda - 4) (\lambda - 6) = 0\)
\(\therefore\) eigenvalues are \(\lambda_{1} = 1, \; \lambda_{2} = 4, \; \lambda_{3} = 6\)
Eigenvactor
\((\lambda I_{n} - A) \vec{v} = \vec{0}\)
Or
\[ \begin{bmatrix} \lambda - 1 & -2 & -3 \\ 0 & \lambda - 4 & -5 \\ 0 & 0 & \lambda - 6 \end{bmatrix} \vec{v} = \vec{0} \]
For \(\lambda_{1} = 1\)
\[ \begin{bmatrix} 1 - 1 & -2 & -3 \\ 0 & 1 - 4 & -5 \\ 0 & 0 & 1 - 6 \end{bmatrix} \vec{v} = \vec{0} \]
\[ \begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix} \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
A1 <- matrix(c(0, -2,-3, 0, -3, -5, 0, 0, -5), nrow = 3, byrow = T)
#Divide row 1 by -2
A1[1,] <- A1[1,] / -2
A1## [,1] [,2] [,3]
## [1,] 0 1 1.5
## [2,] 0 -3 -5.0
## [3,] 0 0 -5.0#Multiply row 1 by 3 and add it to 2 row
A1[2,] <- A1[1,] *3 + A1[2,]
A1## [,1] [,2] [,3]
## [1,] 0 1 1.5
## [2,] 0 0 -0.5
## [3,] 0 0 -5.0#Divide row 2 by -0.5
A1[2,] <- A1[2,] / -.5
A1## [,1] [,2] [,3]
## [1,] 0 1 1.5
## [2,] 0 0 1.0
## [3,] 0 0 -5.0#Multiply row 2 by 1.5 and subtract it from row 1
A1[1,] <- A1[1,] - A1[2,] * 1.5
A1## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 -5#Multiply row 2 by 5 and add it to 3 row
A1[3,] <- A1[3,] + A1[2,] * 5
A1## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0\[ (\lambda I_{3} - A)v = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \end{bmatrix} \]
\(v_{2} = 0\)
\(v_{3} = 0\)
Let, \(v_{1} = t\)
\[ E_{\lambda =1}= t \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \] For t is any real number
For \(\lambda_{2} = 4\)
\[ \begin{bmatrix} 4 -1 & -2 & -3 \\ 0 & 4 - 4 & -5 \\ 0 & 0 & 4 - 6 \end{bmatrix} \vec{v} = \vec{0} \]
\[ \begin{bmatrix} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{bmatrix} \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
A4 <- matrix(c(3, -2, -3, 0, 0, -5, 0, 0, -2), nrow = 3, byrow = T)
#Divide the row 1 by 3
A4[1,] <- A4[1,]/3
A4## [,1] [,2] [,3]
## [1,] 1 -0.6666667 -1
## [2,] 0 0.0000000 -5
## [3,] 0 0.0000000 -2#Divide the row 2 by -5
A4[2,] <- A4[2,] / -5
A4## [,1] [,2] [,3]
## [1,] 1 -0.6666667 -1
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 -2#Multiply 2 row by 1 and add it to 1 row
A4[1,] <- A4[1,] + A4[2,]
A4## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 -2#Multiply row 2 by 2 and add it to row 3
A4[3,] <- A4[2,] * 2 + A4[3,]
A4## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0\[ (\lambda I_{3} - A)v = \begin{bmatrix} 1 & -0.6666667 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \end{bmatrix} \]
\(v_{1} - 0.6666667v_{2} = 0\)
\(v_{3} = 0\)
Let, \(v_{2} =t\)
\(\therefore 0.6666667t = v_{1}\)
\[ E_{\lambda =4}= t \begin{bmatrix} 0.6666667 \\ 1 \\ 0 \end{bmatrix} \]
For t is any real number
For \(\lambda_{3} = 6\)
\[ \begin{bmatrix} 6 -1 & -2 & -3 \\ 0 & 6 - 4 & -5 \\ 0 & 0 & 6 - 6 \end{bmatrix} \vec{v} = \vec{0} \]
\[ \begin{bmatrix} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
A6 <- matrix(c(5, -2, -3, 0, 2, -5, 0, 0, 0), nrow = 3, byrow = T)
#divide the row 1 by 5
A6[1,] <- A6[1,] / 5
A6## [,1] [,2] [,3]
## [1,] 1 -0.4 -0.6
## [2,] 0 2.0 -5.0
## [3,] 0 0.0 0.0#Divide the row 2 by 2
A6[2,] <- A6[2,] / 2
A6## [,1] [,2] [,3]
## [1,] 1 -0.4 -0.6
## [2,] 0 1.0 -2.5
## [3,] 0 0.0 0.0#Multiply row 2 by 0.4 and add it to row 1
A6[1,] <- A6[2,] * .4 + A6[1,]
A6## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0\[ (\lambda I_{3} - A)v = \begin{bmatrix} 1 & -0.4 & -1.6 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \end{bmatrix} \]
\(v_{1} - 1.6v_{3} = 0\)
\(v_{2} - 2.5v_{3}= 0\)
Let, \(v_{3} =t\)
\(\therefore 1.6t = v_{1}\)
\(2.5t = v_{2}\)
\[ E_{\lambda =6}= t \begin{bmatrix} 1.6 \\ 2.5 \\ 0 \end{bmatrix} \]
For t is any real number
Using R
A <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow = 3, byrow = T)
eigen(A)## $values
## [1] 6 4 1
##
## $vectors
## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0