\[ \begin{aligned} \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} \end{aligned} \]
require(Matrix)## Loading required package: MatrixA <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), ncol = 4,byrow = T)
rankMatrix(A)## [1] 4
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 8.881784e-16let A be an mxn matrix. Then the row rank of A is equal to the maximum number of linear independent row vectors of A. Hence the row-rank of mxn matrix is at most m.
Similarly, the column-rank of A is the maximum number of linear independent columns of A. Hence the column-rank of A is at most n.
Since for any matrix A mxn over field F, the row-rank = column-rank. Therefore to be the common value of row-rank A and column-rank A, clearly rank A has to be equal or less of Min (m,n) for any mxn matrix
From the above, for any m x n matrix, if m is greater than n, then the maximum rank of the matrix is m. and if m is less than n, then the maximum rank of the matrix is n.
The rank of a matrix would be zero only if the matrix had no elements. However, If a matrix had even one element, its minimum rank would be one.
\[ \begin{aligned} \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 1 \end{bmatrix} \end{aligned} \]
B <- matrix(c(1, 2, 1, 3, 6, 3, 2, 4, 2), ncol = 3, byrow = T)
rankMatrix(B)## [1] 1
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 6.661338e-16\[ A = \begin{aligned} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \end{aligned} \] \[ \text{ Definition: A scalar} \lambda \text{ is called an eigenvalue of the n × n matrix A is there is a nontrivial solution x of Ax =} \lambda x \\ \text{ . Such an x is called an eigenvector corresponding to the eigenvalue} \lambda.\\ \text{ A scalar} \lambda \text{ is an eigenvalue of an n×n matrix A if and only if} \lambda \text{ satisfies the characteristic equation.}\\ det( A-\lambda I) = 0 \] \[ \lambda I = \begin{aligned} \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} \end{aligned} \] \[ A-\lambda I = \begin{aligned} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} = \begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6 -\lambda \end{bmatrix} = 0 \end{aligned} \] * We can calculate the characteristic polymomial to be:
\[ \begin{aligned} Det(A-\lambda I) &= (\lambda -1)((\lambda -4)(\lambda -6) -0) + 2((0-0) -3(0-0)) = 0\\ & = \lambda^3 - 11\lambda^2 + 34\lambda - 24 \\ &= (\lambda -1)(\lambda -4)(\lambda-6)\\ & = 0 \end{aligned} \] \[ \text{Setting this equal to 0 and solving for} \lambda \text{ we get that } \lambda \text{ = 1, 4,6. These are the three eigenvalues of A.} \]
Now, lets plug in each value of \(\lambda\) into the matrix:
\(( \lambda I_3 - A ) = \left[ {\begin{array}{ccc} 0 & -2 & -3\\ 0 & -3 & -5\\ 0 & 0 & -5\\ \end{array} } \right]\)
A1 <- matrix(c(0,0,0,-2,-3,0,-3,-5,-5),nrow=3)
# eliminate A1[2,3] by subtracting row 3 from row 2
# eliminate A1[1,3] by subtracting 3/5 * row 3 from row
A1[2,] <- A1[2,] - A1[3,]
A1[1,] <- A1[1,] - (3/5)*A1[3,]
A1## [,1] [,2] [,3]
## [1,] 0 -2 0
## [2,] 0 -3 0
## [3,] 0 0 -5# eliminate A1[1,2] by subtracting (2/3)*A1[2,]
A1[1,] <- A1[1,] - (2/3)*A1[2,]
A1## [,1] [,2] [,3]
## [1,] 0 0 0
## [2,] 0 -3 0
## [3,] 0 0 -5# Get matrix into reduced row echelon form by making pivots = 1
# Then move the zero rows to the bottom
A1[2,] <- A1[2,]/-3
A1[3,] <- A1[3,]/-5
A1 <- A1[c(2,3,1),]\(( \lambda I_3 - A )v = \left[ {\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{array} } \right] \left[ {\begin{array}{c} v_1\\ v_2\\ v_3\\ \end{array} } \right]\)
\(v_2 = 0 \\ v_3 = 0\)
Let \(v_1 = t\)
\(E_{\lambda = 1} = t \left[ {\begin{array}{c} 1\\ 0\\ 0\\ \end{array} } \right]\)
where \(t\) is real
\(( \lambda I_3 - A ) = \left[ {\begin{array}{ccc} 3 & -2 & -3\\ 0 & 0 & -5\\ 0 & 0 & -2\\ \end{array} } \right]\)
A4 <- matrix(c(3,0,0,-2,0,0,-3,-5,-2),nrow=3)
# Eliminate A4[2,3] by subtracting (5/2) * row 3 from row 2
# Eliminate A4[1,3] by subtracting (3/2) * row 3 from row 1
A4[2,] <- A4[2,] - (5/2)*A4[3,]
A4[1,] <- A4[1,] - (3/2)*A4[3,]
A4## [,1] [,2] [,3]
## [1,] 3 -2 0
## [2,] 0 0 0
## [3,] 0 0 -2# Divide row 1 by 3 to make the pivot 1
# Divide row 3 by -2 to make the pivot 1
# Move the zero row to the bottom
A4[1,] <- A4[1,]/3
A4[3,] <- A4[3,]/-2
A4 <- A4[c(1,3,2),]\(( \lambda I_3 - A )v = \left[ {\begin{array}{ccc} 1 & -\frac{2}{3} & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{array} } \right] \left[ {\begin{array}{c} v_1\\ v_2\\ v_3\\ \end{array} } \right]\)
\(v_1 -\frac{2}{3}v_2 = 0 \\ v_3 = 0\)
let \(v_2 = t\)
So: \(\frac{2}{3}t = v_1\\\)
\(E_{\lambda = 4} = t \left[ {\begin{array}{c} \frac{2}{3}\\ 1\\ 0\\ \end{array} } \right]\)
Where t is real
A6 <- matrix(c(5,0,0,-2,2,0,-3,-5,0),nrow=3)
# Eliminate A6[1,2] by subtracting -1 * row 2
A6[1,] <- A6[1,] + A6[2,]
# Divide Row 1 by 5 and row 2 by 2 to make the pivots 1
A6[1,] <- A6[1,]/5
A6[2,] <- A6[2,]/2
A6## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0\(( \lambda I_3 - A )v = \left[ {\begin{array}{ccc} 1 & 0 & -1.6\\ 0 & 1 & -2.5\\ 0 & 0 & 0\\ \end{array} } \right] \left[ {\begin{array}{c} v_1\\ v_2\\ v_3\\ \end{array} } \right]\)
\(v_1 - 1.6v_3 = 0 \\ v_2 - 2.5v_3 = 0\)
let \(v_3 = t\)
So
\(V_1 = 1.6t \\ v_2 = 2.5t\)
\(E_{\lambda = 6} = t \left[ {\begin{array}{c} 1.6\\ 2.5\\ 1\\ \end{array} } \right]\)