\[A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \\ \end{bmatrix}\]
In order to find the rank of the matrix A we have to transform the matrix A to its row echelon form and count the number of non-zero rows.
#build the inial matrix
A = matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3),nrow=4, ncol=4, byrow = TRUE)
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3#row2<-row1+row2
A[2,] = A[1,] + A[2,]
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 0 2 4 7
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3#row2<-row2/row1
A[2,] = A[2,]/A[1,]
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3.000000 4.00
## [2,] 0 1 1.333333 1.75
## [3,] 0 1 -2.000000 1.00
## [4,] 5 4 -2.000000 -3.00#row4<-row4-5row1
A[4,] = A[4,] - 5*A[1,]
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3.000000 4.00
## [2,] 0 1 1.333333 1.75
## [3,] 0 1 -2.000000 1.00
## [4,] 0 -6 -17.000000 -23.00#row3<-row3-row2
A[3,] = A[3,] - A[2,]
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3.000000 4.00
## [2,] 0 1 1.333333 1.75
## [3,] 0 0 -3.333333 -0.75
## [4,] 0 -6 -17.000000 -23.00#row4<-6row2+row4
A[4,] = 6*A[2,] + A[4,]
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3.000000 4.00
## [2,] 0 1 1.333333 1.75
## [3,] 0 0 -3.333333 -0.75
## [4,] 0 0 -9.000000 -12.50#row3<-row3/element[3,3]
A[3,] = A[3,]/A[3,3]
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3.000000 4.000
## [2,] 0 1 1.333333 1.750
## [3,] 0 0 1.000000 0.225
## [4,] 0 0 -9.000000 -12.500#row4<-9row3+row4
A[4,] = 9*A[3,] + A[4,]
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3.000000 4.000
## [2,] 0 1 1.333333 1.750
## [3,] 0 0 1.000000 0.225
## [4,] 0 0 0.000000 -10.475The rank of a matrix equals to the number of non-zero rows of row echelon matrix. If matrix has m rows then there are a possibility that all rows of row echelon matrix are non-zero. So that, the maximum rank of a matrix is m.
\[B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \\ \end{bmatrix}\]
In order to find the rank of the matrix B we have to transform the matrix A to its row echelon form and count the number of non-zero rows.
#build the inial matrix
B = matrix(c(1,2,1,3,6,3,2,4,2),nrow=3, ncol=3, byrow = TRUE)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2#row2<-row2-3row1
B[2,] = B[2,] - 3*B[1,]
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 2 4 2#swap row2 and row3
x = B[2,]
B[2,] = B[3,]
B[3,] = x
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 2 4 2
## [3,] 0 0 0#row2<-row2-2row1
B[2,] = B[2,] - 2*B[1,]
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0The number of non-zero rows in row echelon matrix is 1. So that, the rank of matrix B 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.
\[A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{bmatrix}\]
Let’s calculate eigenvalues
\[det( \begin{bmatrix} 1 - {\lambda} & 2 & 3 \\ 0 & 4 - {\lambda} & 5 \\ 0 & 0 & 6 - {\lambda} \\ \end{bmatrix})=0\]
(1 - \({\lambda}\))(4 - \({\lambda}\))(6 - \({\lambda}\)) = 0
The eigenvalues are
\({\lambda}_1\)=1, \({\lambda}_2\)=4, \({\lambda}_3\)=6
Let’s check eigenvalues using the built-in function ‘eigen’
#build the inial matrix
A = matrix(c(1,2,3,0,4,5,0,0,6),nrow=3, ncol=3, byrow = TRUE)
A## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6print(eigen(A)$values)## [1] 6 4 1Now, let’s calculate eigenvectors
\[\left[ \begin{array}{cc} 1-1 & 2 & 3 \\ 0 & 4-1 & 5 \\ 0 & 0 & 6-1 \end{array} \right] % \left[ \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{cc} 0 \\ 0 \\ 0 \end{array} \right]\]
\[\left[ \begin{array}{cc} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{array} \right] % \left[ \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{cc} 0 \\ 0 \\ 0 \end{array} \right]\]
\(2*x_2+3*x_3=0\)
\(3*x_2+5*x_3=0\)
\(5*x_3=0\)
\(x_1\) is a free variable \(x_1=1\)
\(x_3=0\) \(=>\) \(2*x_2+3*0=0\) \(=>\) \(x_2=0\)
\[x_{\lambda=1}=\begin{bmatrix} 1 \\0 \\0\end{bmatrix}\]
\[\left[ \begin{array}{cc} 1-4 & 2 & 3 \\ 0 & 4-4 & 5 \\ 0 & 0 & 6-4 \end{array} \right] % \left[ \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{cc} 0 \\ 0 \\ 0 \end{array} \right]\]
\[\left[ \begin{array}{cc} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{array} \right] % \left[ \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{cc} 0 \\ 0 \\ 0 \end{array} \right]\]
\(-3*x_1+2*x_2+3*x_3=0\)
\(0*x_1+0*x_2+5*x_3=0\)
\(0*x_1+0*x_2+2*x_3=0\)
\(x_2=1\)
\(0*x_1+0*x_2+2*x_3=0\) \(=>\) \(x_3=0\)
\(-3*x_1+2*x_2+3*0=0\) \(=>\) \(3*x_1=2*x_2\) , \(x_1=2*1/3=2/3\)
\[x_{\lambda=1}=\begin{bmatrix} 2/3 \\1 \\0\end{bmatrix}\]
\[\left[ \begin{array}{cc} 1-6 & 2 & 3 \\ 0 & 4-6 & 5 \\ 0 & 0 & 6-6 \end{array} \right] % \left[ \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{cc} 0 \\ 0 \\ 0 \end{array} \right]\]
\[\left[ \begin{array}{cc} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{array} \right] % \left[ \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{cc} 0 \\ 0 \\ 0 \end{array} \right]\]
\(-5*x1+2*x2+3*x3=0\)
\(0*x1-2*x2+5*x3=0\)
\(0*x1+0*x2+0*x3=0\)
\(x_3\) is free variable, \(x_3=1\)
\(0*x_1-2*x_2+5*1=0\) \(=>\) \(-2*x_2+5*1=0\), \(-2*x_2=-5*1\), \(x_2=5/2\)
\(-5*x1+2*5/2+3*1=0\) \(=>\) \(-5*x1+5+3*1=0\) \(=>\) \(x_1=8/5\)
\[x_{\lambda=1}=\begin{bmatrix} 8/5 \\5/2 \\1\end{bmatrix}\]
Now, let’s check eigenvectors using built-in ‘eigen’ function
print(eigen(A)$vectors)## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0