\[\begin{equation*} A = \begin{pmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{pmatrix} \end{equation*}\]
If A is an m x n matrix, then the rank of A is the dimension of the column space.To do that we first row-reduce the matrix since that will help us determine bases for the null space and column space.From below, we see that the RREF matrix is a square identity matrix with 4 rows and 4 columns. Here the number of dependent or “enslaved” columns is 4.D = {1,2,3,4}, For each index in D, theorem BCS creates a single basis vector. In total, the basis will have 4 vectors, so the column space of A will have the dimension 4 and we write r(A) = 4. This is shown below by the r function called matrix.rank which returns the rank of a square numeric matrix.
library(matrixcalc)
library(pracma)## Warning: package 'pracma' was built under R version 3.6.3A <- matrix(data = c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow = 4, ncol = 4, byrow = TRUE); print(A)## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3r_matrix <- rref(A); print(r_matrix)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1ra <- matrix.rank(r_matrix)
sprintf("The rank of matrix A is: %d", ra)## [1] "The rank of matrix A is: 4"Per theorem CRN, if r denotes the number of pivot columns (or non-zero rows), then r(A) = r. As we saw with the example above, if the number of non-zero rows is the maximum number of rows, then the rank equals the number of rows of the mxn matrix. So, the maximum rank r(A) = m. The minimum rank could be 1.
\[\begin{equation*} B = \begin{pmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{pmatrix} \end{equation*}\]
As we did before, reduce matrix B to Reduced-row Echelon Form (RREF). We see it has only one non-zero row from below, hence the rank is 1.This can be found using the R function matrix.rank as we previously employed in problem set 1.
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0## [1] "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 poynomial and show your solution.
\[\begin{equation*} A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix} \end{equation*}\]
Here A is a square matrix of size 3. Then the characteristic polynomial of A is the polynomial p_A(x) defined by \[\begin{equation*} p_A(x) = det(A - x * I_3) \end{equation*}\]
where I_3 is the identity matrix of dimensions 3. Substituting the values of matrix A and x * I_3 into the RHS of the equation, we get below: \[\begin{equation*} p_A(x) = det(\begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix} - \begin{pmatrix} x & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & x \end{pmatrix}) \end{equation*}\]
Simplifying the RHS into one matrix gives us below:
\[\begin{equation*} det(\begin{pmatrix} (1-x) & 2 & 3 \\ 0 & (4-x) & 5 \\ 0 & 0 & (6-x) \end{pmatrix}) = 0 \end{equation*}\]
For a 3x3 or higher-order matrix, we can solve the determinant as below:
\[\begin{equation*} (1 - x) * (\begin{pmatrix} (4-x) & 5 \\ 0 & (6-x) \end{pmatrix}) - 2 * (\begin{pmatrix} 0 & 5 \\ 0 & (6-x) \end{pmatrix}) + 3 * (\begin{pmatrix} 0 & (4-x) \\ 0 & 0 \end{pmatrix}) = 0 \end{equation*}\]
The above equation reduces to a single line below:
\[\begin{equation*} (1 - x) * (((4 - x) * (6 - x)) - (0 * 5)) - 2 * ((0 * (6 - x)) - (0 * 5)) + 3 * ((0 * 0) - (0 * (4 -x))) \end{equation*}\]
which further reduces to the factored form of the polynomial below and this equation is in effect our characteristic polynomial.
\[\begin{equation*} p_A(x) = (1 - x) * ((4 - x) * (6 - x)) \end{equation*}\]
Now for a square matrix A, lambda is an eigenvalue of A if and only if the below condition is satisfied, i.e., Eigenvalues of a matrix are roots of the characteristic polynomial, so we find the roots.
\begin{pmatrix} det(A - I_3) * x = 0\ \end{pmatrix}
Setting our characteristic polynomial equation to 0, we get the following roots -> x = 1, x = 4 and x = 6. By theorem EMRCP, we then get the eigenvalues below.
\[\begin{equation*} (\lambda = 1) \\ (\lambda = 4) \\ (\lambda = 6) \end{equation*}\]
Now, solving for Eigen vectors.
\[\begin{pmatrix} v * (A - I * \lambda) = 0\\ \end{pmatrix}\]When Eigenvalue = 1, \[\begin{equation*} (\begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}) * \begin{pmatrix} v1 \\ v2 \\ v3 \end{pmatrix} = 0 \end{equation*}\]
OR
\[\begin{equation*} (\begin{pmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{pmatrix} * \begin{pmatrix} v1 \\ v2 \\ v3 \end{pmatrix}) = 0 \end{equation*}\]
Written as a system of equations as shown in the weekly reading pdf, we get:
\[\begin{equation*} 5 * v_3 = 0\\ v_3 = 0 \\ 3 * v_2 + 5 * v_3 = 0\\ v_3 = 0 -> v_2 = 0\\ v_1 = 1 \end{equation*}\]
The Eigenvectors for Eigenvalue = 1 are \[\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\]When Eigenvalue = 4, \[\begin{equation*} (\begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix} - \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix}) * \begin{pmatrix} v1 \\ v2 \\ v3 \end{pmatrix} = 0 \end{equation*}\]
OR
\[\begin{equation*} (\begin{pmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{pmatrix} * \begin{pmatrix} v1 \\ v2 \\ v3 \end{pmatrix}) = 0 \end{equation*}\]
Written as a system of equations as shown in the weekly reading pdf, we get:
\[\begin{equation*} 2 * v_3 = 0\\ v_3 = 0 \\ -3 * v_1 + 2 * v_2 = 0\\ v_1 = 2/3 * v_2 \\ v_2 = 1 \end{equation*}\]
The Eigenvectors for Eigenvalue = 4 are \[\begin{pmatrix} 2/3 \\ 1 \\ 0 \end{pmatrix}\]When Eigenvalue = 6, \[\begin{equation*} (\begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix} - \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix}) * \begin{pmatrix} v1 \\ v2 \\ v3 \end{pmatrix} = 0 \end{equation*}\]
OR
\[\begin{equation*} (\begin{pmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{pmatrix} * \begin{pmatrix} v1 \\ v2 \\ v3 \end{pmatrix}) = 0 \end{equation*}\]
Written as a system of equations as shown in the weekly reading pdf, we get:
\[\begin{equation*} (-2 * v_2) + (5 * v_3) = 0\\ v_2 = 5/2 * v_3 \\ -5 * v_1 + 2 * v_2 + 3 * v_3 = 0\\ 5 * v_1 = 8 * v_3 \\ v_1 = 8/5 \end{equation*}\]
The Eigenvectors for Eigenvalue = 6 are \[\begin{pmatrix} 8/5 \\ 5/2 \\ 1 \end{pmatrix}\]The manually derived values above are confirmed through the use of the R function below.
library(pracma)
A <- matrix(c(1,2,3,0,4,5,0,0,6), 3, 3, byrow=TRUE); A## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6I <- diag(3)
rref(A - I)## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0rref(A - 4 * I)## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0rref(A - 6 * I)## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0eigen(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 0