Problem Set 1

One

A <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow=4, byrow=TRUE)
# Consider matrix A
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
# RREF of A
library(pracma)
rref(A)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

There are 4 pivot rows, so the rank of \(A\) is 4. Double-check with R.

# Rank of A
Rank(A)
## [1] 4

Two

Since rank of a matrix is defined as the number of linearly independent column vectors and is equal to the number of linearly independent row vectors, given an \(m\times n\) matrix with \(m>n\), the maximum rank is the lower value \(rank_{max}=n\). Assuming a non-zero matrix, the rank should be at least 1, so \(rank_{min}=1\).

Three

B <- matrix(c(1,2,1,3,6,3,2,4,2), nrow=3, byrow=TRUE)

B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
# Rank of B
Rank(B)
## [1] 1

In addition to getting an answer in R, we can see that row 2 and row 3 are multiples of row 1, so rows 1 and 2 and 1 and 3 are linearly dependent. There is only 1 linearly independent row, so \(rank(B)=1\).

Problem Set 2

A <- matrix(c(1,2,3,0,4,5,0,0,6), nrow=3, byrow=TRUE)
# Consider matrix A
A
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6

Since \(A\) is a triangular matrix, its eigenvalues are values on the diagonal, so \(\lambda_1=1\), \(\lambda_2=4\) and \(\lambda_3=6\).

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

\(det(A− \lambda I) = 0 \Longleftrightarrow det(A− \lambda I)\) = \(det(\) \(\left[ \begin{array}{cccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{array} \right] \\ \) - \(\lambda\) \(\left[ \begin{array}{cccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] \\ \) \()\) \(= det(\) \(\left[ \begin{array}{cccc} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \\ \end{array} \right] \\ \)\()=0\)

using the last row which has 2 zeros, we get:

\(det(A−\lambda I) = 0 \Longleftrightarrow (6-\lambda)[(4-\lambda)(1-\lambda)- (2*0)] = (6-\lambda)(4-\lambda)(1-\lambda) = 0\)

The characteristic polynomial is:

\[f_A(\lambda) = (1-\lambda)(4-\lambda)(6-\lambda) = 24-34\lambda+11\lambda^2-\lambda^3\]

If \(\lambda=1\), then \(A - 1I_3\) is row-reduced to

rref(A - 1 * diag(3))
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

\[ \begin{bmatrix} 0 &1 &0\\ 0 &0 &1\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] Then \(v_1=v_1\) and \(v_2=0\) and \(v_3=0\). The eigenspace is
\[ E_{\lambda=1}= span \Bigg( \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \Bigg) \]

If \(\lambda=4\), then \(A - 4I_3\) is row-reduced to

rref(A - 4 * diag(3))
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0

\[ \begin{bmatrix} 1 &-\frac{2}{3} &0\\ 0 &0 &1\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]

Then \(v_1 - \frac{2}{3}v_2=0\) and \(v_3=0\).

Or \(v_1=v_1\) and \(v_2=\frac{3}{2}v_1=1.5v_1\) and \(v_3=0\).

The eigenspace is

\[ E_{\lambda=4}= span \Bigg( \begin{bmatrix} 1\\ 1.5\\ 0 \end{bmatrix} \Bigg) \]

Finally, if \(\lambda=6\), then \(A - 6I_3\) is row-reduced to

rref(A - 6 * diag(3))
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

\[ \begin{bmatrix} 1 &0 &-1.6\\ 0 &1 &-2.5\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]

Then \(v_1-1.6v_3=0\) and \(v_2-2.5v_3=0\).

Or \(v_1=1.6v_3\) and \(v_2=2.5v_3\) and \(v_3=v_3\).

The eigenspace is

\[ E_{\lambda=6}= span \Bigg( \begin{bmatrix} 1.6\\ 2.5\\ 1 \end{bmatrix} \Bigg) \]