What is the rank of the matrix A?
\[A=\begin{vmatrix} 1 & 2 & 3& 4 \\ -1 & 0 & 1 & 3\\ 0&1&-2&1\\5&4&-2&3\end{vmatrix}\]
The rank of a matrix is defined as the number of linearly independent rows or linearly independent columns in a matrix A. The process for computing the rank involves translating a matrix into echelon form.
We will use the library pracma to reduce the matrix to echelon form. After reduction, we see that the rank of matrix A is 4 since there are 4 non zero linearly independent rows
library(pracma)
A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), nrow = 4,ncol = 4)
rref(A)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1Let us verify using built in R commands
print(qr(A)$rank)## [1] 4Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
Maximum rank
The maximum rank can be full rank with leading entry of 1 in each column in row echelon form, and n is the number of columns. So the maximum rank can be n. The matrix A above is a matrix with full rank.
https://www.youtube.com/watch?v=HZy-sFsCCxI
Minimum rank
The minimum rank is 1 assuming the matrix in non zero. This can happen when the rest of the rows are linearly dependant. The matrix X below is such a matrix.
X<-matrix(c(1,2,3,4,2,4,6,8,3,6,9,12), nrow=3, byrow=T)
X## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 2 4 6 8
## [3,] 3 6 9 12print("The rank of this matrix is")## [1] "The rank of this matrix is"print(qr(X)$rank)## [1] 1What is the rank of matrix B?
B <- matrix(c(1,3,2,2,6,4,1,3,2), nrow = 3,ncol = 3)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2The rows are linearly dependant, so it can be easily guessed that the rank is 1. Let us test using R
print(rref(B))## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0print(qr(B)$rank)## [1] 1Compute the Eigenvalues and Eigenvectors of matrix A. You’ll need to show your work. You’ll need to write the characteristic polynomial and show your solution.
A <- matrix(c(1,0,0,2,4,0,3,5,6), nrow = 3, ncol = 3)
A## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6The matrix A is in the form of an upper triangular matrix.
The determinant is therefore 146 which is equal to 24. Let us verify
print(det(A))## [1] 24det(A-λI)=0
\[det(\begin{vmatrix} 1-λ & 2 & 3\\ 0 & 4-λ & 5\\ 0 &0 & 6-λ\end{vmatrix})=0\]
The characteristic polynomial is therefore
\[(1-λ)(4-λ)(6-λ)=0\]
The eigenvalues are λ=1, λ=4, λ=6
Let us confirm using R
e<-eigen(A)
e$values## [1] 6 4 1For eigenvectors, we have to solve (A-λnI)Xn=0
When λ=1
a1<-matrix(c(0,0,0,2,3,0,3,5,5), nrow =3, byrow=F)
print("the matrix is")## [1] "the matrix is"a1## [,1] [,2] [,3]
## [1,] 0 2 3
## [2,] 0 3 5
## [3,] 0 0 5print("Reduced form")## [1] "Reduced form"print(rref(a1))## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0\[\begin{vmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 &0 & 0\end{vmatrix} \begin{vmatrix}X1\\X2\\X3\end{vmatrix} = \begin{vmatrix}0\\0\\0\end{vmatrix}\]
X1 is free, X2=0, X3=0 so the eigenvector for λ=1 is \[X1=\begin{vmatrix}1\\0\\0\end{vmatrix}\]
When λ=4,
a2<-matrix(c(-3,0,0,2,0,0,3,5,2), nrow =3, byrow=F)
print("the matrix is")## [1] "the matrix is"a2## [,1] [,2] [,3]
## [1,] -3 2 3
## [2,] 0 0 5
## [3,] 0 0 2print("Reduced form")## [1] "Reduced form"print(rref(a2))## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0Keeping X2 free, X3=0, X1-0.6667X2=0, so the eigenvector for λ=4 is
\[X2=\begin{vmatrix}1\\3/2\\0\end{vmatrix}\]
Similarly for λ=4
a3 <- matrix(c(-5,0,0,2,-2,0,3,5,0), nrow =3, byrow=F)
print("the matrix is")## [1] "the matrix is"a3## [,1] [,2] [,3]
## [1,] -5 2 3
## [2,] 0 -2 5
## [3,] 0 0 0print("Reduced form")## [1] "Reduced form"print(rref(a3))## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0X3 is free, X1-1.6X3=0, X2-2.5X3=0 X1=1.6, X2=2.5, so the eigenvector for λ=6 is
\[X3=\begin{vmatrix}1.6\\2.5\\1\end{vmatrix}\]
Let us check using R,
print(eigen(A))## eigen() decomposition
## $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 0If we consider X3[3,1]=1, X3[1,1]=0.510/0,319 = 1.6 similarly, X3[2,1] becomes 2.5, so my solution looks ok.
print(0.5108407/0.3192754)## [1] 1.6print(0.7981886/0.3192754)## [1] 2.5