Problem set 1

(1) What is the rank of the matrix A?

A <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), 4, 4, byrow=TRUE)
qr(A)$rank
## [1] 4

(2) Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

Answer:
Maximum rank = n (the smaller of the row or column matrix)
Minimum rank = 1 (since the matrix is non-zero)

(3) What is the rank of matrix B?

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

Problem set 2

Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work. You’ll need to write out the characteristic polynomial and show your solution. \[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right]\]

Eigenvalues

det(A- \(\lambda\) \(I_{3}\)) = 0 :

\[\left[\begin{array} {rrr} \lambda - 1 & -2 & -3 \\ 0 & \lambda - 4 & -5 \\ 0 & 0 & \lambda - 6 \end{array}\right] = 0\]

(1- \(\lambda\))(4- \(\lambda\))(6- \(\lambda\)) = 0

Eigenvalues = 1, 4, and 6.

The chractoeristic polynomial is p(\(\lambda\)) = (1- \(\lambda\))(4- \(\lambda\))(6- \(\lambda\)) = \(-\lambda^{3}\) \(-11\lambda^{2}\)+ \(34\lambda\)-24

Eigenvectors

1). \(\lambda\) = 1

\[\left[\begin{array} {rrr} 1 - 1 & -2 & -3 \\ 0 & 1 - 4 & -5 \\ 0 & 0 & 1 - 6 \end{array}\right] = \left[\begin{array} {rrr} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{array}\right]\]

row reduction: \[\left[\begin{array} {rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right]\]

Eigenspace: \[\mathbf{E_{\lambda = 1}(A)}= \left(\begin{array} {rrr} 1 \\ 0\\ 0 \end{array}\right)\]

2). \(\lambda\) = 4

\[\left[\begin{array} {rrr} 4 - 1 & -2 & -3 \\ 0 & 4 - 4 & -5 \\ 0 & 0 & 4 - 6 \end{array}\right] = \left[\begin{array} {rrr} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{array}\right]\]

row reduction: \[\left[\begin{array} {rrr} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right]\]

Eigenspace: \[\mathbf{E_{\lambda = 4}(A)}= \left(\begin{array} {rrr} 1 \\ -2/3\\ 0 \end{array}\right)\]

3). \(\lambda\) = 6

\[\left[\begin{array} {rrr} 6 - 1 & -2 & -3 \\ 0 & 6 - 4 & -5 \\ 0 & 0 & 6 - 6 \end{array}\right] = \left[\begin{array} {rrr} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{array}\right]\]

row reduction: \[\left[\begin{array} {rrr} 1 & 0 & -1.6 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{array}\right]\]

Eigenspace: \[\mathbf{E_{\lambda = 6}(A)}= \left(\begin{array} {rrr} 1.6 \\ 2.5\\ 1 \end{array}\right)\]