\(A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & 3 \end{bmatrix}\)
The rank of the above 4 X 4 matrix is 4 due to the presence of 4 non-zero pivots, as shown below using the row reduced echelon form :
(A <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3),nrow=4, byrow=TRUE))## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3#Eliminate the element (2,1) ==> r1 + r2
(E21 <- matrix(c(1,0,0,0,1,1,0,0,0,0,1,0,0,0,0,1), nrow=4, byrow=TRUE))## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 1 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1(A <- E21 %*% A)## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 0 2 4 7
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3#Eliminate the element (4,1) ==> r4 - 5 r1
(E41 <- matrix(c(1,0,0,0,0,1,0,0,0,0,1,0,-5,0,0,1), nrow=4, byrow=TRUE))## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] -5 0 0 1(A <- E41 %*% A)## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 0 2 4 7
## [3,] 0 1 -2 1
## [4,] 0 -6 -17 -23#Eliminate the element (3,2) ==> r3 - r2/2
(E32 <- matrix(c(1,0,0,0,0,1,0,0,0,-0.5,1,0,0,0,0,1), nrow=4, byrow=TRUE))## [,1] [,2] [,3] [,4]
## [1,] 1 0.0 0 0
## [2,] 0 1.0 0 0
## [3,] 0 -0.5 1 0
## [4,] 0 0.0 0 1(A <- E32 %*% A)## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.0
## [2,] 0 2 4 7.0
## [3,] 0 0 -4 -2.5
## [4,] 0 -6 -17 -23.0#Eliminate the element (4,2) ==> r4 + 3 r2
(E42 <- matrix(c(1,0,0,0,0,1,0,0,0,0,1,0,0,3,0,1), nrow=4, byrow=TRUE))## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 3 0 1(A <- E42 %*% A)## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.0
## [2,] 0 2 4 7.0
## [3,] 0 0 -4 -2.5
## [4,] 0 0 -5 -2.0#Eliminate the element (4,3) ==> r4 - (5 * r3) / 4
(E43 <- matrix(c(1,0,0,0,0,1,0,0,0,0,1,0,0,0,-1.25,1), nrow=4, byrow=TRUE))## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.00 0
## [2,] 0 1 0.00 0
## [3,] 0 0 1.00 0
## [4,] 0 0 -1.25 1(A <- E43 %*% A)## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.000
## [2,] 0 2 4 7.000
## [3,] 0 0 -4 -2.500
## [4,] 0 0 0 1.125#Conclusion : There is no row with all zeroes, and there are 4 non-zero pivots so, the rank of the matrix is 4.The maximum rank of an \(m\) x \(n\) matrix will be the smaller of the 2 dimensions, so, \(m > n\) , then the max rank is \(n\).
If a matrix is non-zero, then obviosly it must contain atleast 1 non-zero pivot, hence the minimum rank is 1.
\(B=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix}\)
As given below, the matrix \(B\) has only one non-zero row , hence, its rank is 1.
(B <- matrix(c(1,2,1,3,6,3,2,4,2),nrow=3, byrow=TRUE))## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2#Eliminate the element (2,1) ==> r2 - 3r1
(E21 <- matrix(c(1,0,0,-3,1,0,0,0,1), nrow=3, byrow=TRUE))## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] -3 1 0
## [3,] 0 0 1(B <- E21 %*% B)## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 2 4 2#Eliminate the element (3,1) ==> r3 - 2r1
(E31 <- matrix(c(1,0,0,0,1,0,-2,0,1), nrow=3, byrow=TRUE))## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] -2 0 1(B <- E31 %*% B)## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0#Conclusion: 2 rows are all of zeroes, and only 1 non-zero row present, so the Rank(B) = 1Compute the eigenvalues and eigenvectors for matrix \(A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}\)
\(\lambda\) is an eigenvalue of \(A\) iff \(det(A - \lambda I_n) = 0\) ==> \[ det \left(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) = 0 \] ==> \[ det \left( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} \right) = 0 \] ==> \[ det \left( \begin{bmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6 - \lambda \end{bmatrix} \right) = 0 \] ==> \[ (1 - \lambda ) * det \begin{bmatrix} 4 - \lambda & -5 \\ 0 & 6 - \lambda \end{bmatrix} - 2 * det \begin{bmatrix} 0 & -5 \\ 0 & 6 - \lambda \end{bmatrix} + 3 * det \begin{bmatrix} 0 & 4 - \lambda \\ 0 & 0 \end{bmatrix} = 0 \] ==> \[ (1 - \lambda)((4- \lambda)(6 - \lambda) - 0) + 2(0 - 0) + 3(0 - 0) = 0 \] So, the characteristic equation is: \[ \lambda^3 - 11\lambda^2 + 34\lambda - 24 = (\lambda - 1)(\lambda - 4)(\lambda - 6) = 0 \] hence , the eigenvalues are: \(\lambda = 1\), \(\lambda = 4\), and \(\lambda = 6\).
Using the below equation, lets calculate the eigenvectors for the eigenvalues : \[ (A - \lambda I_n) \vec{v} = 0 \]
\[ \left( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0 \]
From the third row, \(5 v_3 = 0\) so \(v_3 = 0\). Substituting it in row 2 equation, we will get \(3 v_2 = 0\) , so \(v_2 = 0\). Finally from the first row, \(0 v_1 = 0\). Using the above information gives the following eigenvector when \(\lambda = 1\) , ( because we need to make sure the vector is non-zero vector ) : \[ v= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \]
\[ \left( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \right) \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0 \] from third row, \(2 v_3 = 0\) or, \(v_3 = 0\)
substituting this in the first row gives the following: \[ -3 v_1 + 2 v_2 = 0 ==> \frac{3}{2} v_1 = v_2 \]
Hence, the eigenvector when \(\lambda = 4\) is: \[ v= \begin{pmatrix} 1 \\ \frac{3}{2} \\ 0 \end{pmatrix} \]
\[ \left( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \right) \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0 \]
So, the equations are: \[-5 v_1 + 2 v_2 + 3 v_3 = 0 , -2 v_2 + 5 v_3 = 0 \] by substituting, we will get \[ v_3 = \frac{2}{5} v_2 , v_2 = \frac{25}{16} v_1 \] by substituting \(v_2\) in the first equation , we will get: \[ v_3 = \frac{5}{8} v_1 \]
Hence, the eigenvector when \(\lambda = 6\) is: \[ v= \begin{pmatrix} 1 \\ \frac{25}{16} \\ \frac{5}{8} \end{pmatrix} \]
(A <- matrix(c(1,2,3,0,4,5,0,0,6), nrow=3, byrow=TRUE))## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6(e <- eigen(A))## $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