Q1

## [1] 4

  1. The maximum rank of a matrix where m > n cannot be larger than the min(m, n). The minimum rank of any matrix outside of a zero matrix is 1.

## [1] 1

Q2

## [1] 6 4 1
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0

The characteristic polynomial is: \(p_{A}(\lambda) = (1 - \lambda)(4 - \lambda)(6 - \lambda)\)

The eigenspace when \(\lambda = 1\) is:

##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

\[ \left[\begin{array} {rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array} {rrr} v_{1} \\ v_{2} \\ v_{3} \end{array}\right] = \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array}\right] \]

Therefore

\[ E_{\lambda=1} = \left[\begin{array} {rrr} 1 \\ 0 \\ 0 \end{array}\right] \]

The eigenspace when \(\lambda = 4\) is:

##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0

\[ \left[\begin{array} {rrr} 1 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array} {rrr} v_{1} \\ v_{2} \\ v_{3} \end{array}\right] = \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array}\right] \]

Therefore

\[ E_{\lambda=4} = \left[\begin{array} {rrr} 1 \\ 1.5 \\ 0 \end{array}\right] \]

The eigenspace when \(\lambda = 6\) is:

##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

\[ \left[\begin{array} {rrr} 1 & 0 & -1.6 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array} {rrr} v_{1} \\ v_{2} \\ v_{3} \end{array}\right] = \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array}\right] \]

Therefore

\[ E_{\lambda=6} = \left[\begin{array} {rrr} 1.6 \\ 2.5 \\ 1 \end{array}\right] \]