Exercise LT.C16

Find the matrix representation of T: \(\mathbb{C}^3 \rightarrow \mathbb{C}^4, T\left \{\begin{array}{r} x \\ y \\ z \\ \end{array} \right \} = \left \{\begin{array}{r} 3x+2y+z \\ x+y+z \\ x-3y \\ 2x+3y+z\\ \end{array} \right \}\).

Matrix Representation:

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

Exercise LT.C30

Define the linear transformation

\[T: \mathbb{C}^3 \rightarrow \mathbb{C}^2, T\left[\begin{array}{r} x_1 \\ x_2 \\ x_3 \\ \end{array} \right] = \left[\begin{array}{r} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3\\ \end{array} \right]\]

Computer the preimages, \(T^{-1}\left[\begin{array}{r} 2 \\ 3 \\ \end{array} \right]\) and \(T^{-1}\left[\begin{array}{r} 4 \\ -8 \\ \end{array} \right]\)

# For T^(-1) [2, 3]
# Will create an augmented matrix with the linear transformation
require(pracma)
## Loading required package: pracma
first <- matrix(c(2, -1, 5, 2, -4, 2, -10, 3), nrow = 2, byrow = TRUE)
print("Augmented Matrix for T^-1 [2,3]: ")
## [1] "Augmented Matrix for T^-1 [2,3]: "
first
##      [,1] [,2] [,3] [,4]
## [1,]    2   -1    5    2
## [2,]   -4    2  -10    3
print("Matrix in RREF: ")
## [1] "Matrix in RREF: "
rref(first)
##      [,1] [,2] [,3] [,4]
## [1,]    1 -0.5  2.5    0
## [2,]    0  0.0  0.0    1
# For T^(-1) [4, -8]
second <- matrix(c(2, -1, 5, 4, -4, 2, -10, -8), nrow = 2, byrow = TRUE)
print("Augmented Matrix for T^-1 [4,-8]: ")
## [1] "Augmented Matrix for T^-1 [4,-8]: "
second
##      [,1] [,2] [,3] [,4]
## [1,]    2   -1    5    4
## [2,]   -4    2  -10   -8
print("Matrix in RREF: ")
## [1] "Matrix in RREF: "
rref(second)
##      [,1] [,2] [,3] [,4]
## [1,]    1 -0.5  2.5    2
## [2,]    0  0.0  0.0    0

Regarding \(T^{-1}\left[\begin{array}{r} 2 \\ 3 \\ \end{array} \right]\), given that this system is inconsistent (as there is a nonzero column in the last column), there is no vector that is a member of \(T^{-1}\).

Regarding \(T^{-1}\left[\begin{array}{r} 4 \\ -8 \\ \end{array} \right]\), there are two free variables, \(x_2\) and \(x_3\). If we set these free variables to equal zero (arbitrarily), we can create a solution for these preimages. \(x_1 = 2 + (1/2)x_2 + (-5/2)x_3\)

\[T^{-1}\left[\begin{array}{r} 4 \\ -8 \\ \end{array} \right] = \left[\begin{array}{r} 2 \\ 0 \\ 0 \\ \end{array} \right] + x_2\left[\begin{array}{r} 1/2 \\ 1 \\ 0 \\ \end{array} \right] + x_3\left[\begin{array}{r} -5/2\\ 0 \\ 1 \\ \end{array} \right]\]