Question 1

Consider a matrix A:

\(A=\begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3\end{bmatrix}\)

So:

A = matrix(c(1, 2, 3, 4, -1, 0, 1, 3, 0, 1, -2, 1, 5, 4, -2, -3), ncol = 4, nrow = 4, byrow = TRUE)

qr(A)$rank
## [1] 4

Question 2

Maximum rank is n columns, since M > N and rank A <= N

Maximum Rank: Suppose following non-zero 5 X 4 matrix:

$A= \[\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}\]

$

A has to be at least 1 if non-zero

Question 3

What is the rank of the matrix:

$B= \[\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2\end{bmatrix}\]

$

B = matrix(c(1, 2, 1, 3, 6, 3, 2, 4, 2), nrow = 3, ncol = 3, byrow = TRUE)
qr(B)$rank
## [1] 1

Problem Set 2

A <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow = 3, byrow = T)

\(\lambda I_3=\begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}\)

\(\lambda I_3-A=\begin{bmatrix} \lambda -1 & 0 & 0 \\ 0 & \lambda -4 & 0 \\ 0 & 0 & \lambda -6 \end{bmatrix}\)

(λ−1)[(λ−4)(λ−6)−0]−2(0)+3(0)=0

(λ−1)(λ^2−10λ+24)=0

λ3−11λ2+34λ−24=0 => Characteristic polynomial

(λ-1)(λ-4)(λ-6)=0

λ = 1,4,6 = Eigenvalues

λ = 1

\(A<-\begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5\end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)

A1 <- matrix(c(0, -2,-3, 0, -3, -5, 0, 0, -5), nrow = 3, byrow = T)
#Row reduction 
(A1[1,] <- A1[1,] / -2)
## [1] 0.0 1.0 1.5
(A1[2,] <- A1[1,] *3 + A1[2,])
## [1]  0.0  0.0 -0.5
(A1[2,] <- A1[2,] / -.5)
## [1] 0 0 1
(A1[1,] <- A1[1,] - A1[2,] * 1.5)
## [1] 0 1 0
(A1[3,] <- A1[3,] + A1[2,] * 5)
## [1] 0 0 0
A1
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

\((\lambda I_3 - A)v = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}\)

\(E_\lambda=1=t \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\)

For $ _2=4$:

A2 <- matrix(c(3, -2, -3, 0, 0, -5, 0, 0, -2), nrow = 3, byrow = T)
#Row reduction
(A2[1,] <- A2[1,]/3)
## [1]  1.0000000 -0.6666667 -1.0000000
(A2[2,] <- A2[2,] / -5)
## [1] 0 0 1
(A2[1,] <- A2[1,] + A2[2,]) 
## [1]  1.0000000 -0.6666667  0.0000000
(A2[3,] <- A2[2,] * 2 + A2[3,])
## [1] 0 0 0

\[(\lambda I_3 - A)v = \begin{bmatrix} 1 & -.6666667 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}\]

\(E_\lambda =_4=t \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\)

For λ3=6

A3 <- matrix(c(5, -2, -3, 0, 2, -5, 0, 0, 0), nrow = 3, byrow = T)
#Row reduction
(A3[1,] <- A3[1,] / 5)
## [1]  1.0 -0.4 -0.6
(A3[2,] <- A3[2,] / 2)
## [1]  0.0  1.0 -2.5
(A3[1,] <- A3[2,] * .4 + A3[1,])
## [1]  1.0  0.0 -1.6
A3
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

\[(\lambda I_3 - A)v = \begin{bmatrix} 1 & -.4 & 0 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0\end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}\]

\(2.5t=v_2\)

\(E_\lambda=_6=t \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\)