require("pracma")## Loading required package: pracmaProblem 1. (1)What is the rank of the matrix A?
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 -3#Change a Matrix Into its Echelon Form
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 1The maximum number of linearly independent vectors in a matrix is equal to the number of non-zero rows in its row echelon matrix. Therefore, to find the rank of a matrix, we simply transform the matrix to its row echelon form and count the number of non-zero rows.
There are 4 pivot rows, so the rank of \(A\) is 4.
Rank(A)## [1] 4double check by r programing which shows Rank=4.
reference: https://stattrek.com/matrix-algebra/echelon-transform.aspx#MatrixA https://stattrek.com/matrix-algebra/matrix-rank.aspx
(2)Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
Anwser: maximum rank is n, minum rank is 1.
reference: https://stattrek.com/matrix-algebra/matrix-rank.aspx
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 2#change matrix B into echelon form
rref(B)## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0Accordint to the echelon form of matrix B, we conclude the rank of B is 1.
#Double check with R programming
Rank(B)## [1] 1Problem 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 <- 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 6Because \(A\) is a “upper triangular matrix”“, its eigenvalues are values on the diagonal, so \(\lambda_1=1\), \(\lambda_2=4\) and \(\lambda_3=6\).
reference:http://mathonline.wikidot.com/determining-eigenvalues-from-upper-triangular-matrices-of-li https://math.stackexchange.com/questions/1971598/eigenvectors-of-a-triangular-matrix
# Double-check eigenvalues in R programming
eigen(A)$values## [1] 6 4 1The “characteristic polynomial”" is \(p_A(\lambda) = (1-\lambda)(4-\lambda)(6-\lambda)\) or \(p_A(\lambda) = -\lambda^3+11\lambda^2-34\lambda+24\).
If \(\lambda=1\), then \(A - 1I_3\) is row-reduced to
rref(A - 1 * diag(3))## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0\[ \begin{bmatrix} 0 &1 &0\\ 0 &0 &1\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]
Then \(v_1=v_1\) and \(v_2=0\) and \(v_3=0\). The “eigenspace” is
\[
E_{\lambda=1}=
\Bigg\langle \Bigg\{
\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}
\Bigg\} \Bigg \rangle
\]
If \(\lambda=4\), then \(A - 4I_3\) is row-reduced to
rref(A - 4 * diag(3))## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0\[ \begin{bmatrix} 1 &-\frac{2}{3} &0\\ 0 &0 &1\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]
Then \(v_1 - \frac{2}{3}v_2=0\) and \(v_3=0\).
Or \(v_1=v_1\) and \(v_2=\frac{3}{2}v_1=1.5v_1\) and \(v_3=0\).
The “eigenspace” is
\[ E_{\lambda=4}= \Bigg\langle \Bigg\{ \begin{bmatrix} 1\\ 1.5\\ 0 \end{bmatrix} \Bigg\} \Bigg \rangle \] Finally, if \(\lambda=6\), then \(A - 6I_3\) is row-reduced to
rref(A - 6 * diag(3))## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0\[ \begin{bmatrix} 1 &0 &-1.6\\ 0 &1 &-2.5\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]
Then \(v_1-1.6v_3=0\) and \(v_2-2.5v_3=0\).
Or \(v_1=1.6v_3\) and \(v_2=2.5v_3\) and \(v_3=v_3\).
The “eigenspace” is
\[ E_{\lambda=6}= \Bigg\langle \Bigg\{ \begin{bmatrix} 1.6\\ 2.5\\ 1 \end{bmatrix} \Bigg\} \Bigg \rangle \]