The rank of a matrix is defined as (a) the maximum number of linearly independent column vectors in the matrix or (b) the maximum number of linearly independent row vectors in the matrix.
\[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array}\right] \]
Transform matrix A to reduced echelon form, which is:
\[\mathbf{A} = \left[\begin{array} {rrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]
Looking at the resultant matrix, the rank is 4.
Confirm using R to calculate rank:
A <- matrix(c(1, 2, 3, 4, -1, 0, 1, 3, 0, 1, -2, 1, 5,4, -2, -3), nrow= 4, byrow = T)
y <- qr(A); y$rank ## [1] 4The rank of matrix A = 4
The maximum rank can be no greater than n.
The 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 transform the matrix to its row echelon form and count the number of non-zero rows
Assuming a non-zero matrix, the minimum rank would be 1.
An example would be:
\[\mathbf{B} = \left[\begin{array} {rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}\right] \]
\[\mathbf{B} = \left[\begin{array} {rrr} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \\ \end{array}\right] \]
Transform matrix B to reduced echelon form, which is:
\[\mathbf{B} = \left[\begin{array} {rrr} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}\right] \]
B <- matrix(c(1, 2, 1, 3, 6, 3, 2, 4, 2), nrow= 3, byrow = T)
y <- qr(B); y$rank ## [1] 1The rank of matrix B = 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 polynomial and show your solution.
\[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{array}\right] \]
\(A\)x = \(\lambda\)x
det( \(A\) - \(\lambda\) \(I\) ) = 0
\[ A - \lambda I = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{pmatrix} - \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \\ \end{pmatrix} \]
\[ A - \lambda I = \begin{vmatrix} 1- \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6- \lambda \\ \end{vmatrix} \]
\[ = (1 - \lambda) \begin{vmatrix} 4 - \lambda & 5\\ 0 & 6- \lambda \\ \end{vmatrix} - (2) \begin{vmatrix} 0 & 5 \\ 0 & 6- \lambda \\ \end{vmatrix} + (3) \begin{vmatrix} 0 & 4 - \lambda \\ 0 & 0 \\ \end{vmatrix} \]
\[ = (1 - \lambda) \begin{pmatrix} (4 - \lambda )(6 - \lambda ) - 5(0) \end{pmatrix} - (2) \begin{pmatrix} (0) (6 - \lambda ) - 5(0) \end{pmatrix} + 3 \begin{pmatrix} (0)(0) - (4 - \lambda )(0) \end{pmatrix} \]
Simplifying this:
\[ = (1 - \lambda) \begin{pmatrix} \lambda^2 - 10\lambda + 24 \end{pmatrix} \]
We can calculate the characteristic polymomial to be:
\[ \mathbf{p(\lambda)} = -\lambda^3 + 11\lambda^2 - 34\lambda + 24 = 0 \]
\[ \mathbf{p(\lambda)} = (-1)(-\lambda^3 + 11\lambda^2 - 34\lambda + 24) = 0(-1) \]
\[ \mathbf{p(\lambda)} = \lambda^3 - 11\lambda^2 + 34\lambda - 24 = 0 \]
\[ \mathbf{p(\lambda)} = (\lambda - 1)(\lambda - 4)(\lambda - 6) = 0 \]
Using the characteristic polynomial, we determine that the eigenvalues of \(A\), given by \(\lambda\), are 1, 4, 6.
\(\lambda_1\) = 1 \(\lambda_2\) = 4 \(\lambda_3\) = 6
Using \(\lambda_1\) = 1:
\[ A - \lambda I = \begin{pmatrix} 1- 1 & 2 & 3 \\ 0 & 4 - 1 & 5 \\ 0 & 0 & 6 - 1 \\ \end{pmatrix} = \begin{pmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \\ \end{pmatrix} \]
Using Gaussian Elimination on the resulting augmented matrix:
\[ \begin{pmatrix} 0 & 2 & 3 & 0\\ 0 & 3 & 5 & 0\\ 0 & 0 & 5 & 0\\ \end{pmatrix} \]
\[ R_1 = 1/2(R_1) = \begin{pmatrix} 0 & 1 & 3/2 & 0\\ 0 & 3 & 5 & 0\\ 0 & 0 & 5 & 0\\ \end{pmatrix} \]
\[ R_2 = -3(R_1) = \begin{pmatrix} 0 & 1 & 3/2 & 0\\ 0 & 0 & 1/2 & 0\\ 0 & 0 & 5 & 0\\ \end{pmatrix} \]
\[ R_2 = 2(R_2) = \begin{pmatrix} 0 & 1 & 3/2 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 5 & 0\\ \end{pmatrix} \]
\[ R_3 = -5(R_2) = \begin{pmatrix} 0 & 1 & 3/2 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} \]
\[ R_1 = R_1 -3/2(R_2) = \begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} \]
Solving the system:
\begin{align} x_1 & is & a & free & variable\\ x_2 &=0\\ x_3 &=0\\ \end{align}\[ v_1 = \begin{pmatrix} x_1 \\ 0 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} \]
Using \(\lambda_2\) = 4:
\[ A - \lambda I = \begin{pmatrix} 1- 4 & 2 & 3 \\ 0 & 4 - 4 & 5 \\ 0 & 0 & 6 - 4 \\ \end{pmatrix} = \begin{pmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \\ \end{pmatrix} \]
Using Gaussian Elimination on the resulting augmented matrix:
\[ \begin{pmatrix} -3 & 2 & 3 & 0\\ 0 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 \end{pmatrix} \]
\[ R_1 = -1/3(R_1) = \begin{pmatrix} 1 & -2/3 & -1 & 0\\ 0 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 \end{pmatrix} \]
\[ R_2 = (R_1)/5 = \begin{pmatrix} 1 & -2/3 & -1 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0 \end{pmatrix} \]
\[ R_3 = R_3 - 3(R_2) = \begin{pmatrix} 1 & -2/3 & -1 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
\[ R_1 = R_1 + R_2 = \begin{pmatrix} 1 & -2/3 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
Solving the system:
\begin{align} x_1 -2/3x_2 = 0; & x_1 = 2/3x_2 \\ x_3 & = 0\\ \end{align}\[ v_2 = \begin{pmatrix} 2/3x_2 \\ x_2 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 0 \\ \end{pmatrix} \]
Using \(\lambda_3\) = 6:
\[ A - \lambda I = \begin{pmatrix} 1 -6 & 2 & 3 \\ 0 & 4 -6 & 5 \\ 0 & 0 & 6 -6 \\ \end{pmatrix} = \begin{pmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \\ \end{pmatrix} \]
Using Gaussian Elimination on the resulting augmented matrix:
\[ \begin{pmatrix} -5 & 2 & 3 & 0\\ 0 & -2 & 5 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} \]
\[ R_1 = -1/5(R_1) = \begin{pmatrix} 1 & -2/5 & -3/5 & 0\\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0\\ \end{pmatrix} \]
\[ R_2 = -1/2(R_2) = \begin{pmatrix} 1 & -2/5 & -3/5 & 0\\ 0 & 1 & -5/2 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
\[ R_1 = R_1 -2/5(R_2) = \begin{pmatrix} 1 & 0 & -8/5 & 0\\ 0 & 1 & -5/2 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
Solving the system:
\[ \begin{align} x_1 + 0x_2 -8/5x_3 &=& 0\\ 0x_1 + x_2 -5/2x_3 &=& 0 \end{align} \]
\[ \begin{align} x_1 &=& 8/5x_3 \\ x_2 &=& 5/2x_3 \end{align} \]
\[ v_3 = \begin{pmatrix} 8/5x_3 \\ 5/2x_3 \\ 10 \\ \end{pmatrix} = \begin{pmatrix} 16 \\ 25 \\ 10 \\ \end{pmatrix} \]