IS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS - 2014
- Problem set 1
- What is the rank of the matrix A?
\[ A = \begin{bmatrix} 1&2&3&4\\ −1&0&1&3\\ 0&1&−2&1\\ 5&4&−2&−3 \end{bmatrix} \]
Convert A to RREF
\[ A = \begin{bmatrix} 1&2&3&4\\ −1&0&1&3\\ 0&1&−2&1\\ 5&4&−2&−3 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix} \]
Rank is 4. There are no linear independent rows
A = matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3),nrow=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 -3pracma::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 1Matrix::rankMatrix(A)## [1] 4
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 8.881784e-16- Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
The maximum rank would be m as it is possible as in problem set one that no rows are linearly independent. The minimum rank is 1
- What is the rank of matrix B? \[ B = \begin{bmatrix} 1&2&1\\ 3&6&3\\ 2&4&2 \end{bmatrix} \]
Convert B to reduced row echelon form. Count number of non-zero rows
\[ B = \begin{bmatrix} 1&2&1\\ 3&6&3\\ 2&4&2 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1&2&1\\ 0&0&0\\ 0&0&0 \end{bmatrix} \]
B = matrix(c(1,2,1,3,6,3,2,4,2),nrow=3,byrow=TRUE)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2pracma::rref(B)## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0Matrix::rankMatrix(B)## [1] 1
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 6.661338e-16- 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.
\[ A = \begin{bmatrix} 1&2&3\\ 0&4&5\\ 0&0&6 \end{bmatrix} \]
\[ \lambda I_3 = \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} \]
\[ \lambda I_3 = - A = \begin{bmatrix} \lambda-1 & -2 & -3 \\ 0 & \lambda-4 & -5 \\ 0 & 0 & \lambda-6 \end{bmatrix} \]
Apply rule of Sarrus
\[ \lambda I_3 = - A = \begin{bmatrix} \lambda-1 & -2 & -3 \\ 0 & \lambda-4 & -5 \\ 0 & 0 & \lambda-6 \end{bmatrix} \begin{matrix} \lambda-1 & -2 \\ 0 & \lambda-4 \\ 0 & 0 \end{matrix} \]
\[ (\lambda-1 )(\lambda-4)(\lambda-6) + (-2)(-5)(0) + (-3)(0)(0) - (-2)(0)(\lambda-6) - (\lambda-1)(-5)(0) - (-3)(\lambda-4)(0) \] Everything drops out except the first terms \[ (\lambda-1 )(\lambda-4)(\lambda-6) \] This leaves us with three eigenvalues \[\lambda = 1\] \[\lambda = 4\] \[\lambda = 6\]
Eigenvector for \[\lambda = 1\]
Plug value in for lambda
\[ \lambda I_3 = - A = \begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1\\v_2\\v_3 \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix} \\ v_1 = t\\ v_2 = 0\\ v_3 = 0 \\ E_1 = t \begin{bmatrix}1\\0\\0\end{bmatrix} \]
Eigenvector for \[\lambda = 4\]
Plug value in for lambda
\[ \lambda I_3 = - A = \begin{bmatrix} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & 2 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1 & 2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1\\v_2\\v_3 \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix} \\ v_1 + 2/3v_2 = 0 \\ v_3 = 0 \]
Eigenvector for \[\lambda = 6\]
Plug value in for lambda
\[ \lambda I_3 = - A = \begin{bmatrix} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1 & 0 & -1.6 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1\\v_2\\v_3 \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix} \\ v_1 - 1.6v_3 = 0 \rightarrow 0.625v_1 = v_3\\ v_2 - 2.5v_3 = 0 \rightarrow 0.4v_2 = v_3 \\ v_3 = t \\ E_6 = t \begin{bmatrix} 0.625\\0.4\\0\end{bmatrix} \]
A = matrix(c(1,2,3,0,4,5,0,0,6),nrow=3,byrow=TRUE)
A## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6eigen(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 0B = matrix(c(0,-2,-3,0,-3,-5,0,0,-5),nrow=3,byrow=TRUE)
B## [,1] [,2] [,3]
## [1,] 0 -2 -3
## [2,] 0 -3 -5
## [3,] 0 0 -5pracma::rref(B)## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0C = matrix(c(3,-2,-3,0,0,-5,0,0,0),nrow=3,byrow=TRUE)
C## [,1] [,2] [,3]
## [1,] 3 -2 -3
## [2,] 0 0 -5
## [3,] 0 0 0pracma::rref(C)## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0D = matrix(c(5,-2,-3,0,2,-5,0,0,0),nrow=3,byrow=TRUE)
D## [,1] [,2] [,3]
## [1,] 5 -2 -3
## [2,] 0 2 -5
## [3,] 0 0 0pracma::rref(D)## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0