library(pracma)

A: (1)

The set of pivot columns of any reduced row echelon form matrix is known as Rank.

#Create a matrix
A = matrix(c(1,-1,0,5, 2,0,1,4, 3,1,-2,-2, 4,3,1,-3),ncol = 4)

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
#Using qr function calculate rank
rankA <- qr(A)
rankA$rank
## [1] 4

Rank of the matrix A: 4

#reduced row echelon form matrix
rrefA = rref(A)
rrefA
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

Reduced row echelon form of matrix A shows each column is linearly indepentent, hence rank is 4.

A: (2)

According to theorem Computing Rank and Nullity(CRN), suppose that A is an m x n matrix and B is a row-equivalent matrix in reduced row-echelon form. Let r denote the number of pivot columns (or the number of nonzero rows). Then \(r(A) = r\) and \(n(A) = n - r\).

According to theorem Rank Plus Nullity(RPNC) is Columns, suppose that A is an m x n matrix. Then \(r(A) + n(A) = n\).

Using both theorems, it can be concluded that an m x n matrix can only have single rank and it is defined as minimum(m,n). If rank is equal to n(number of columns) \(r(A) = n\), then it is called as full rank. A matrix is said to be rank deficient if it does not have full rank, \(r(A) < n\).

Using given conditions of the problem, m > n, hence \(r(A) \le n\).

A: (3)

#Create a matrix
B = matrix(c(1,3,2, 2,6,4, 1,3,2),ncol = 3)

B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
#Using qr function calculate rank
rankB <- qr(B)
rankB$rank
## [1] 1

Rank of the matrix B: 1

#reduced row echelon form matrix
rrefB = rref(B)
rrefB
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0

Reduced row echelon form of matrix B shows only one non-zero row, hence rank is 1.

A: Characteristic polynomial

#define matrix
A = matrix(c(1,0,0, 2,4,0, 3,5,6),ncol = 3)

cp = charpoly(A)

#Characteristic polynomial
cp
## [1]   1 -11  34 -24

Characteristic polynomial equation for given matrix is \(1x^3 -11x^2 + 34x -24\)

Eigenvalues

According to theorem EMRCP, Eigenvalues of a Matrix are Roots of Characteristic Polynomials Suppose A is a square matrix. Then \(\lambda\) is an eigenvalue of A if and only if \(p_{A} (\lambda) = 0\)

By setting the characteristic polynomial equation equal to zero, \(p_{A}(x) = 1x^3 -11x^2 + 34x -24 = 0\)

ev = eigen(A)
ev$values
## [1] 6 4 1

Roots are \(x = 6, x = 4\) and \(x = 1\).

Eigenvalues of given matrix are \(\lambda = 6, \lambda = 4\) and \(\lambda = 1\)

Eigenvectors

\(\lambda = 6, A - \lambda I_{3}\)

lambda = ev$values[1]
lambdaI = lambda * diag(3)

result = A - lambdaI

#using pracma library get reduced row echelon form for result
rrefMx = rref(result)
rrefMx
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

Null space, \({\cal E}_{A}(6) = {\cal N}(A - (6I_{3}))\)

\({\cal N}(A + 6I_{3}) = \left[\begin{array}{cc}1 & 0 & -1.6 \\0 & 1 & -2.5 \\0 & 0 & 0 \end{array}\right] \left[\begin{array}{cc}v_1 \\ v_2 \\ v_3\end{array}\right] = \left[\begin{array}{cc}0 \\ 0 \\ 0\end{array}\right]\)

Solving the equations, \(v_1 -1.6v_3 = 0,\ v_2 -2.5v_3 = 0\)

\(if\ v_3 = t, then\ v_1 = 1.6t, v_2 = 2.5t\)

\({\cal E}_{A}(6) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = t\left[\begin{array}{cc} 1.6 \\ 2.5 \\ 1\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = 6\) is \({\cal E}_{A}(6) = \left\langle\left[\begin{array}{cc} 1.6 \\ 2.5 \\ 1\end{array}\right]\right\rangle\)

\(\lambda = 4, A - \lambda I_{3}\)

lambda = ev$values[2]
lambdaI = lambda * diag(3)

result = A - lambdaI

#using pracma library get reduced row echelon form for result
rrefMx = rref(result)
rrefMx
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0

Null space, \({\cal E}_{A}(4) = {\cal N}(A - (4I_{3}))\)

\({\cal N}(A + 4I_{3}) = \left[\begin{array}{cc}1 & -0.6666667 & 0 \\0 & 0 & 1 \\0 & 0 & 0 \end{array}\right] \left[\begin{array}{cc}v_1 \\ v_2 \\ v_3\end{array}\right] = \left[\begin{array}{cc}0 \\ 0 \\ 0\end{array}\right]\)

Solving the equations, \(v_1 -0.6666667v_2 = 0,\ v_3 = 0\)

\(if\ v_2 = t, then\ v_1 = 0.6666667t\)

\({\cal E}_{A}(4) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = t\left[\begin{array}{cc} 0.6666667 \\ 1 \\ 0\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Let \(t = 0.6666667 | {t\ \epsilon\ Real\ Numbers}\)

\({\cal E}_{A}(4) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = 0.6666667\left[\begin{array}{cc} 1 \\ 1.5 \\ 0\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = 4\) is \({\cal E}_{A}(4) = \left\langle\left[\begin{array}{cc} 1 \\ 1.5 \\ 0\end{array}\right]\right\rangle\)

\(\lambda = 1, A - \lambda I_{3}\)

lambda = ev$values[3]
lambdaI = lambda * diag(3)

result = A - lambdaI

#using pracma library get reduced row echelon form for result
rrefMx = rref(result)
rrefMx
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

Null space, \({\cal E}_{A}(1) = {\cal N}(A - (1I_{3}))\)

\({\cal N}(A + 1I_{3}) = \left[\begin{array}{cc}0 & 1 & 0 \\0 & 0 & 1 \\0 & 0 & 0 \end{array}\right] \left[\begin{array}{cc}v_1 \\ v_2 \\ v_3\end{array}\right] = \left[\begin{array}{cc}0 \\ 0 \\ 0\end{array}\right]\)

Solving the equations, \(v_2 = 0,\ v_3 = 0\)

\(let\ v_1 = t\)

\({\cal E}_{A}(1) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = t\left[\begin{array}{cc} 1 \\ 0 \\ 0\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = 1\) is \({\cal E}_{A}(1) = \left\langle\left[\begin{array}{cc} 1 \\ 0 \\ 0\end{array}\right]\right\rangle\)

Solving using eigen function from R

#eigenvectors values from eigen function
ev$vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0

Eigenspace for \(\lambda = 6\)

#for lambda to 6
ev$values[1]
## [1] 6
es = ev$vectors[,1]
es
## [1] 0.5108407 0.7981886 0.3192754
#let minT be min value of vector
minT = min(es)

\({\cal E}_{A}(6) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = t\left[\begin{array}{cc} 0.5108407 \\ 0.7981886 \\ 0.3192754\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Let \(t = 0.3192754 | {t\ \epsilon\ Real\ Numbers}\)

\({\cal E}_{A}(6) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = 0.3192754\left[\begin{array}{cc} 1.6 \\ 2.5 \\ 1\end{array}\right] | 0.3192754\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = 6\) is \({\cal E}_{A}(6) = \left\langle\left[\begin{array}{cc} 1.6 \\ 2.5 \\ 1\end{array}\right]\right\rangle\)

Eigenspace for \(\lambda = 4\)

#for lambda to 4
ev$values[2]
## [1] 4
es = ev$vectors[,2]
es
## [1] 0.5547002 0.8320503 0.0000000
#let minT be min value of vector
minT = min(es[es > 0])

\({\cal E}_{A}(4) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = t\left[\begin{array}{cc} 0.5547002 \\ 0.8320503 \\ 0\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Let \(t = 0.5547002 | {t\ \epsilon\ Real\ Numbers}\)

\({\cal E}_{A}(4) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = 0.5547002\left[\begin{array}{cc} 1 \\ 1.5 \\ 0\end{array}\right] | 0.5547002\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = 4\) is \({\cal E}_{A}(4) = \left\langle\left[\begin{array}{cc} 1 \\ 1.5 \\ 0\end{array}\right]\right\rangle\)

Eigenspace for \(\lambda = 1\)

#for lambda to 1
ev$values[3]
## [1] 1
es = ev$vectors[,3]
es
## [1] 1 0 0
#let minT be min value of vector
minT = min(es[es > 0])

\({\cal E}_{A}(1) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = t\left[\begin{array}{cc} 1 \\ 0 \\ 0\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Let \(t = 1 | {t\ \epsilon\ Real\ Numbers}\)

\({\cal E}_{A}(1) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3\end{array}\right] = 1\left[\begin{array}{cc} 1 \\ 0 \\ 0\end{array}\right] | 1\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = 1\) is \({\cal E}_{A}(1) = \left\langle\left[\begin{array}{cc} 1 \\ 0 \\ 0\end{array}\right]\right\rangle\)

Eigenspace values are same, solving manually matches R function eigen.