\[ \left[ \begin{matrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 &-3 \end{matrix} \right] \]
x <- matrix(c(1, -1, 0, 5, 2, 0, 1, 4, 3, 1, -2, -2, 4, 3, 1, -3), 4, 4)
det(x)## [1] -9The determinate of this matrix is equal to -9. Since the determinant is a non-zero values and the matrix being calculated is a square matrix, we can infer that all rows or columns are linearly independent from one another. Therefore, the matrix is full rank. The rank of this matrix is a 4.
The maximum rank of a given m x n matrix where m > n the maximum rank of that matrix will be n because the rank of a given matrix cannot be larger than the smallest dimension of that matrix. The minimum rank of that matrix, assuming that the matrix is non-zero, would be 1.
\[ \left[ \begin{matrix} 1 & 2 & 1 \\ 3 & 6 & 3\\ 2 & 4 & 2 \end{matrix} \right] \]
y <- matrix(c(1, 3, 2, 2, 6, 4, 1, 3, 2), 3, 3)
det(y)## [1] 0The determinate of this square matrix is equal to 0. That tells me that at least one of the rows or columns are linearly dependent on the other. On further examination, row 3 is 2 times row 1. As well as row two is row 1 plus row 3. Therefore, the rank of this matrix is 1.
\[ \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 4 & 5\\ 0 & 0 & 6 \end{matrix} \right] \] To solve for the eigenvalues, first we must solve for the characteristic polynomials. To solve for the characteristic polynomials, the following equation is used.
\[ p_a(x) = det(A-xI_n) \] Here, I is the identity matrix and where are finding the determinate of the matrix A – x of the identity matrix. \[ p_a(x) = det \left[ \begin{matrix} 1-x & 2 & 3 \\ 0 & 4-x & 5\\ 0 & 0 & 6 - x \end{matrix} \right] \] You can reduce the row by taking out 1-x then solving for the 2x2 matrix.
\[ p_a(x) = det (1-x) \left[ \begin{matrix} 4-x & 5\\ 0 & 6 - x \end{matrix} \right] \] \[ p_a(x) = det(1-x)(4-x)(6-x) - 0 \] Here the characteristic polynomials are as follows: \[ p_a(x) = (1-x)(4-x)(6-x) \]
To solve for the eigenvalues of matrix A Characteristic Polynomials must equal to zero.
\[ p_a(\lambda) = 0 \] \[ p_a(\lambda) = (1-\lambda)(4-\lambda)(6-\lambda)=0 \] Set all the characteristic polynomials to zero and here are the following eigenvalues.
\[ \lambda = 1 \quad \lambda = 4 \quad \lambda = 6 \] To solve for eigenvectors, we first need to solve for the eigenspace of matrix A. The equation to solve for eigenspaces is as follows.
\[ \epsilon(\lambda)=[A-\lambda I_n|0] \] Now we substitute in the lambdas as we solve for the eigenspaces. \[ \lambda = 1 \]
\[ \left[ \begin{matrix} 1-1 & 2 & 3 \\ 0 & 4-1 & 5\\ 0 & 0 & 6 -1 \end{matrix} \right]|0 \]
\[ \left[ \begin{matrix} 0 & 2 & 3 \\ 0 & 3 & 5\\ 0 & 0 & 5 \end{matrix} \right]|0 \]
Here we can make x1 = some scalar value. We will us the scalar value a.Now we solve for x3 by solving the system of equations.
\[ 5x_3 = 0 \quad \rightarrow \quad x_3 = 0 \]
Now we solve for x2
\[ 3x_2 + 5x_3 = 0 \quad \rightarrow \quad 3x_2 + 5(0) = 0 \quad \rightarrow \quad 3x_2 = 0 \quad \rightarrow \quad x_2 = 0 \]
\[ \epsilon(\lambda) = X = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix}= \]
\[ \begin{bmatrix} a \\ 0 \\ 0 \\ \end{bmatrix} = \]
\[ a \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \]
This solve for our eigenspace, and the eigenvector is the vector multiped by the scalar value above. Now we move on to solve lamda = 4.
\[ \lambda = 4 \]
\[ \left[ \begin{matrix} 1-4 & 2 & 3 \\ 0 & 4-4 & 5\\ 0 & 0 & 6 - 4 \end{matrix} \right]|0 \]
\[ \left[ \begin{matrix} -3 & 2 & 3 \\ 0 & 0 & 5\\ 0 & 0 & 2 \end{matrix} \right]|0 \]
Here we can see x3 = 0. Now we solve for x2 by solving the system of equations and make x1 = some scalar value.
\[ -3x_1 +2x_2 = 0 \quad \rightarrow \quad 3x_1 =2 x_2 \quad \rightarrow \quad x_2 = (3/2)x_1 \]
\[ \epsilon(\lambda) = X = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix}= \]
\[ \begin{bmatrix} a \\ a(3/2) \\ 0 \\ \end{bmatrix} = \]
\[ a \begin{bmatrix} 1 \\ 3/2 \\ 0 \\ \end{bmatrix} \]
This solve for our eigenspace, and the eigenvector is the vector multiped by the scalar value above. Now we move on to solve lamda = 6.
\[ \lambda = 6 \]
\[ \left[ \begin{matrix} 1-6 & 2 & 3 \\ 0 & 4-6 & 5\\ 0 & 0 & 6 - 6 \end{matrix} \right]|0 \]
\[ \left[ \begin{matrix} -5 & 2 & 3 \\ 0 & -2 & 5\\ 0 & 0 & 0 \end{matrix} \right]|0 \]
Here we make x2 = some scalar value.Now we solve for x3
\[ -2x_2 +5x_3 = 0 \quad \rightarrow \quad 2x_2 =5 x_3 \quad \rightarrow \quad x_3 = (2/5)x_2 \]
Now we solve for x1
\[ -5x_1 + 2x_2 +3x_3 = 0 \quad \rightarrow \quad -5x_1 + 2x_2 +3(2/5)x_2 = 0 \quad \rightarrow \quad -5x_1+(16/5)x_2 = 0 \quad \rightarrow \quad 5x_1=(16/5)x_2 \quad \rightarrow \quad \] \[ x_1 = (16/25)x_2 \]
\[ \epsilon(\lambda) = X = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix}= \]
\[ \begin{bmatrix} a(16/25) \\ a \\ a (2/5) \\ \end{bmatrix} = \]
\[ a \begin{bmatrix} (16/25) \\ 1 \\ (2/5) \\ \end{bmatrix} \]
Therefore, our eigenvalues are:
\[ \lambda = 1 \quad \lambda = 4 \quad \lambda = 6 \]
And our eigenvectors are:
\[ \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \] \[ \begin{bmatrix} 1 \\ 3/2 \\ 0 \\ \end{bmatrix} \] \[ \begin{bmatrix} (16/25) \\ 1 \\ (2/5) \\ \end{bmatrix} \]