Problem set 1
A = \(\left[\begin{array}{ccc} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3\\ \end{array}\right]\)
Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
What is the rank of matrix B?
B = \(\left[\begin{array}{ccc} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2\\ \end{array}\right]\)
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 = \(\left[\begin{array}{ccc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{array}\right]\)
Please show your work using an R-markdown document.
What is the rank of the matrix A?
A = \(\left[\begin{array}{ccc} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3\\ \end{array}\right]\)
First, we find the row echelon form of matrix A.
Step 1
\(\left[\begin{array}{cccc} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\ 5 & 4 & -2 & -3 \end{array}\right]\underrightarrow{R_{2}=R_{1}+R_{2}}\left[\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7\\ 5 & 4 & -2 & -3 \end{array}\right]\)
Step 2
\(-5R_{1}=\left[\begin{array}{cccc} -5 & -10 & -15 & -20\end{array}\right]\)
\(\left[\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7\\ 5 & 4 & -2 & -3 \end{array}\right]\underrightarrow{R_{3}=-5R_{1}+R_{3}}\left[\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7\\ 0 & -6 & -17 & -23 \end{array}\right]\)
Step 3
\(3R_{2}=\left[\begin{array}{cccc} 0 & 6 & 12 & 21\end{array}\right]\)
\(\left[\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7\\ 0 & -6 & -17 & -23 \end{array}\right]\underrightarrow{R_{3}=3R_{2}+R_{3}}\left[\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7\\ 0 & 0 & -5 & -2 \end{array}\right]\)
Since there are three nonzero rows in the reduced the matrix A, the rank of Matrix A is 3.
A <- matrix(c(1, -1, 5, 2, 0, 4, 3, 1, -2, 4, 3, -3),nrow = 3)
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 5 4 -2 -3library(Matrix)
rankMatrix(A)## [1] 3
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 8.881784e-16Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
According to Stat Trek, “the rank of a matrix is defined as (a) the maximum number of linearly independent column vectors in the matrix” If m > n, the maximum rank of the matrix is n. If the matrix does not contain any elements, the rank of the matrix is 0. If the matrix contains at least 1 element, the minimum rank of the matrix is 0.
What is the rank of matrix B?
B = \(\left[\begin{array}{ccc} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2\\ \end{array}\right]\)
We have a 3x3 matrix. The maximum feasible rank of the matrix is the minimum of the dimension, which is 3. Since column 1 and column 3 are the same, one of them has to go. Column 2 is a multiple of column 1. Row 2 and 3 are each multiples of row 1. The rank of matrix B is 1.
Step 1
\(-3R_{1}=\left[\begin{array}{ccc} -3 & -6 & -3\end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2 \end{array}\right]\underrightarrow{R_{2}=-3R_{1}+R_{2}}\left[\begin{array}{ccc} 1 & 2 & 1\\ 0 & 0 & 0\\ 2 & 4 & 2 \end{array}\right]\)
Step 2
\(-2R_{1}=\left[\begin{array}{ccc} -2 & -4 & -2\end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & 2 & 1\\ 0 & 0 & 0\\ 2 & 4 & 2 \end{array}\right]\underrightarrow{R_{3}=-2R_{1}+R_{3}}\left[\begin{array}{ccc} 1 & 2 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right]\)
Since there are one nonzero rows in the reduced the matrix B, the rank of Matrix B is 1.
B <- matrix(c(1, 3, 2, 2, 6, 4, 1, 3, 2),nrow = 3)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2The rank of matrix B is 1.
rankMatrix(B)## [1] 1
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 6.661338e-16Compute 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 = \(\left[\begin{array}{ccc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{array}\right]\)
Please show your work using an R-markdown document.
\(A(\lambda I-A)-0\)
\(det(\lambda I-A)=0\) \(\lambda I=\lambda\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda \end{array}\right]\)
\(det\left(\left[\begin{array}{ccc} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda \end{array}\right]-\left[\begin{array}{ccc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \end{array}\right]\right)=0\)
\(det\left(\left[\begin{array}{ccc} \lambda-1 & -2 & -3\\ 0 & \lambda-4 & -5\\ 0 & 0 & \lambda-6 \end{array}\right]\right)=0\)
Row 3 has the most zeros. Expand on row 3.
\(0\left|\begin{array}{cc} -2 & -3\\ \lambda-4 & -5 \end{array}\right|-0\left|\begin{array}{cc} \lambda-1 & -3\\ 0 & -5 \end{array}\right|+(\lambda-6)\left|\begin{array}{cc} \lambda-1 & -2\\ 0 & \lambda-4 \end{array}\right|=0\)
\([0]-[0]+(\lambda-6)[(\lambda-1)(\lambda-4)-(0)(-2)]=0\)
\((\lambda-6)[(\lambda-1)(\lambda-4)-0]=0\)
\((\lambda-6)(\lambda-1)(\lambda-4)=0\)
\(\lambda=\) 1, 4, or 6
A <- matrix(c(1, 0, 0, 2, 4, 0, 3, 5, 6), nrow = 3 )
A## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6eigen_A <- eigen(A)
eigen_A$values## [1] 6 4 1Expand the following equation.
\((\lambda-6)(\lambda-1)(\lambda-4)=0\)
\((\lambda^{2}-7\lambda+6)(\lambda-4)=0\)
\((\lambda^{3}-7\lambda^{2}+6\lambda)-(4\lambda^{2}-28\lambda+24)=0\)
\(\lambda^{3}-11\lambda^{2}+34\lambda-24=0\)
Check with R.
library(pracma)##
## Attaching package: 'pracma'## The following objects are masked from 'package:Matrix':
##
## expm, lu, tril, triucharpoly(A)## [1] 1 -11 34 -24\(\lambda=1\)
\(\left[\begin{array}{ccc} \lambda-1 & -2 & -3\\ 0 & \lambda-4 & -5\\ 0 & 0 & \lambda-6 \end{array}\right]=\left[\begin{array}{ccc} 1-1 & -2 & -3\\ 0 & 1-4 & -5\\ 0 & 0 & 1-6 \end{array}\right]=\left[\begin{array}{ccc} 0 & -2 & -3\\ 0 & -3 & -5\\ 0 & 0 & -5 \end{array}\right]\)
We find the reduced echelon form of the matrix above.
\(\left[\begin{array}{ccc} 0 & -2 & -3\\ 0 & -3 & -5\\ 0 & 0 & -5 \end{array}\right]\underrightarrow{R_{1}=-\frac{R_{1}}{2}}\left[\begin{array}{ccc} 0 & 1 & \frac{3}{2}\\ 0 & -3 & -5\\ 0 & 0 & -5 \end{array}\right]\)
\(\left[\begin{array}{ccc} 0 & 1 & \frac{3}{2}\\ 0 & -3 & -5\\ 0 & 0 & -5 \end{array}\right]\underrightarrow{R_{2}=3R_{1}+R_{2}}\left[\begin{array}{ccc} 0 & 1 & \frac{3}{2}\\ 0 & 0 & -\frac{1}{2}\\ 0 & 0 & -5 \end{array}\right]\)
\(\left[\begin{array}{ccc} 0 & 1 & \frac{3}{2}\\ 0 & 0 & -\frac{1}{2}\\ 0 & 0 & -5 \end{array}\right]\underrightarrow{R_{2}=-2R_{2}}\left[\begin{array}{ccc} 0 & 1 & \frac{3}{2}\\ 0 & 0 & 1\\ 0 & 0 & -5 \end{array}\right]\)
\(\left[\begin{array}{ccc} 0 & 1 & \frac{3}{2}\\ 0 & 0 & 1\\ 0 & 0 & -5 \end{array}\right]\underrightarrow{R_{3}=5R_{2}+R_{3}}\left[\begin{array}{ccc} 0 & 1 & \frac{3}{2}\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right]\)
\(\left[\begin{array}{ccc} 0 & 1 & \frac{3}{2}\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right]\underrightarrow{R_{3}=\frac{3}{2}R_{2}+R_{1}}\left[\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right]\)
To find the null space, we solve \(\left[\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right]\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array}\right]=\left[\begin{array}{c} 0\\ 0\\ 0 \end{array}\right]\)
We have \(v_{2}=0;v_{3}=0\)
\(v_{\lambda=1}=\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]\)
\(\lambda=4\)
\(\left[\begin{array}{ccc} \lambda-1 & -2 & -3\\ 0 & \lambda-4 & -5\\ 0 & 0 & \lambda-6 \end{array}\right]=\left[\begin{array}{ccc} 4-1 & -2 & -3\\ 0 & 4-4 & -5\\ 0 & 0 & 4-6 \end{array}\right]=\left[\begin{array}{ccc} 3 & -2 & -3\\ 0 & 0 & -5\\ 0 & 0 & -2 \end{array}\right]\)
We find the reduced echelon form of the matrix above.
\(\left[\begin{array}{ccc} 3 & -2 & -3\\ 0 & 0 & -5\\ 0 & 0 & -2 \end{array}\right]\underrightarrow{R_{1}=\frac{R_{1}}{3}}\left[\begin{array}{ccc} 1 & -\frac{2}{3} & -1\\ 0 & 0 & -5\\ 0 & 0 & -2 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & -\frac{2}{3} & -1\\ 0 & 0 & -5\\ 0 & 0 & -2 \end{array}\right]\underrightarrow{R_{2}=-\frac{R_{2}}{5}}\left[\begin{array}{ccc} 1 & -\frac{2}{3} & -1\\ 0 & 0 & 1\\ 0 & 0 & -2 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & -\frac{2}{3} & -1\\ 0 & 0 & 1\\ 0 & 0 & -2 \end{array}\right]\underrightarrow{R_{1}=R_{2}+R_{1}}\left[\begin{array}{ccc} 1 & -\frac{2}{3} & 0\\ 0 & 0 & 1\\ 0 & 0 & -2 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & -\frac{2}{3} & 0\\ 0 & 0 & 1\\ 0 & 0 & -2 \end{array}\right]\underrightarrow{R_{3}=2R_{2}+R_{3}}\left[\begin{array}{ccc} 1 & -\frac{2}{3} & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & -\frac{2}{3} & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right]\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array}\right]=\left[\begin{array}{c} 0\\ 0\\ 0 \end{array}\right]\)
\(v_{1}-\frac{2}{3}v_{2}=0\Longrightarrow v_{1}=\frac{2}{3}v_{2}\)
\(v_{3}=0\)
\(v_{\lambda=4}=\left[\begin{array}{c} \frac{2}{3}\\ 1\\ 0 \end{array}\right]\)
\(\lambda=6\)
\(\left[\begin{array}{ccc} \lambda-1 & -2 & -3\\ 0 & \lambda-4 & -5\\ 0 & 0 & \lambda-6 \end{array}\right]=\left[\begin{array}{ccc} 6-1 & -2 & -3\\ 0 & 6-4 & -5\\ 0 & 0 & 6-6 \end{array}\right]=\left[\begin{array}{ccc} 5 & -2 & -3\\ 0 & 2 & -5\\ 0 & 0 & 0 \end{array}\right]\)
We find the reduced echelon form of the matrix above.
\(\left[\begin{array}{ccc} 5 & -2 & -3\\ 0 & 2 & -5\\ 0 & 0 & 0 \end{array}\right]\underrightarrow{R_{1}=\frac{R_{1}}{5}}\left[\begin{array}{ccc} 1 & -\frac{2}{5} & -\frac{3}{5}\\ 0 & 2 & -5\\ 0 & 0 & 0 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & -\frac{2}{5} & -\frac{3}{5}\\ 0 & 2 & -5\\ 0 & 0 & 0 \end{array}\right]\underrightarrow{R_{2}=\frac{1}{2}R_{2}}\left[\begin{array}{ccc} 1 & -\frac{2}{5} & -\frac{3}{5}\\ 0 & 1 & -\frac{5}{2}\\ 0 & 0 & 0 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & -\frac{2}{5} & -\frac{3}{5}\\ 0 & 1 & -\frac{5}{2}\\ 0 & 0 & 0 \end{array}\right]\underrightarrow{R_{1}=-\frac{2}{5}R_{2}+R_{1}}\left[\begin{array}{ccc} 1 & 0 & -\frac{8}{5}\\ 0 & 1 & -\frac{5}{2}\\ 0 & 0 & 0 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1 & 0 & -\frac{8}{5}\\ 0 & 1 & -\frac{5}{2}\\ 0 & 0 & 0 \end{array}\right]\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array}\right]=\left[\begin{array}{c} 0\\ 0\\ 0 \end{array}\right]\)
\(v_{1}-\frac{8}{5}v_{3}=0\Longrightarrow v_{1}=\frac{8}{5}v_{3}\)
\(v_{2}-\frac{5}{2}v_{3}=0\Longrightarrow v_{2}=\frac{5}{2}v_{3}\)
\(v_{\lambda=6}=\left[\begin{array}{c} \frac{8}{5}\\ \frac{5}{2}\\ 1 \end{array}\right]\)
The eigenvectors does not match my calculations. Perhaps it is normalized.
eigen_A$vectors## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0Normalized eigenvectors = \(\frac{\left[\begin{array}{c} v_{1}\\ v_{2}\\ \vdots\\ v_{n} \end{array}\right]}{\sqrt{v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}}}\)
eigen_A$vectors## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0norm_eigen_lambda1 <- matrix(c(1,0,0),nrow=3)/sqrt(1^2+0^2+0^2)
norm_eigen_lambda1## [,1]
## [1,] 1
## [2,] 0
## [3,] 0norm_eigen_lambda4 <- matrix(c(2/3,1,0),nrow=3)/sqrt((2/3)^2+1^2+0^2)
norm_eigen_lambda4## [,1]
## [1,] 0.5547002
## [2,] 0.8320503
## [3,] 0.0000000norm_eigen_lambda6 <- matrix(c(8/5,5/2,1),nrow=3)/sqrt((8/5)^2+(5/2)^2+1^2)
norm_eigen_lambda6## [,1]
## [1,] 0.5108407
## [2,] 0.7981886
## [3,] 0.3192754norm_eigenvector <-cbind(norm_eigen_lambda6, norm_eigen_lambda4, norm_eigen_lambda1)
norm_eigenvector## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0Now we check if the our calculated normalized eigenvector matches with the output of the R code. The first element in column one and column to do not match. However, when displaying and comparing the eigenvectors, it appears to match.
norm_eigenvector == eigen_A$vectors## [,1] [,2] [,3]
## [1,] FALSE FALSE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUEnorm_eigenvector## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0eigen_A$vectors## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0norm_eigen_lambda0 <- matrix(c(-1,0,1),nrow=3)/sqrt(1^2+(-1)^2+0^2)
norm_eigen_lambda0## [,1]
## [1,] -0.7071068
## [2,] 0.0000000
## [3,] 0.7071068Upon researching, R does return the normalized eigenvector.