## 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

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