library(pracma)

A:

Kernal of a Linear Transformation(KLT): According to A First Course in Linear Algebra by Robert Beezer, for a linear transformation \(T\colon U \to V\), the kernal is a subset of the domain \(U\). It is similar to null space(\(\cal N\)) of matrix. In set notation,

Suppose \(T\colon U \to V\) is a linear transformation. Then the kernal of \(T\) is the set

\({\cal K}(T)\ =\ \{u\ \epsilon\ U\ |\ T(u)\ =\ 0\}\)

In order to determine elements of \({\mathbb C}^3\) in \({\cal K}(T)\), we need to find vectors \(u\) such that \(T(u) = 0\)

\(T(u) = 0\)

\(\left[\begin{array}{cc}2x+y+z \\x-y+2z \\x+2y-z \end{array}\right] = \left[\begin{array}{cc}0 \\ 0 \\ 0\end{array}\right]\)

Vector equality leads to homogeneous system of 3 equations

\(2x\ +\ y\ + z = 0\)

\(x\ -y\ +2z = 0\)

\(x\ + 2y\ -z = 0\)

Convert into matrix form and change to reduced row echelon form. Then calculate null space(\({\cal N}\))

\(\left[\begin{array}{cc}2 & 1 & 1 \\1 & -1 & 2\\1 & 2 & -1\end{array}\right]\)

#initialize the matrix
A = matrix(c(2,1,1, 1,-1,2, 1,2,-1), nrow=3, byrow=TRUE)

#Make matrix similar to problem
A = t(A)

A
##      [,1] [,2] [,3]
## [1,]    2    1    1
## [2,]    1   -1    2
## [3,]    1    2   -1
#reduced row echelon form
rrefMx = rref(A)
rrefMx
##      [,1] [,2] [,3]
## [1,]    1    0    1
## [2,]    0    1   -1
## [3,]    0    0    0

\(null\ space({\cal N}) = \left[\begin{array}{cc}1 & 0 & 1\\0 & 1 & -1\\0 & 0 & 0\end{array}\right] \left[\begin{array}{cc}x \\ y \\ z\end{array}\right] = \left[\begin{array}{cc}0 \\ 0 \\ 0\end{array}\right]\)

Solving the equations,

\(x +z =0,\ y-z =0\)

\(x = -z,\ y =z\)

if \(z = 1\) then null space \({\cal N}\), is all vectors where \(x =-z\) and \(y = z\).

Thus, \({\cal N} = \left\langle\left[\begin{array}{cc} -1 \\ 1 \\ 1\end{array}\right] \right\rangle\)

Since null space(\({\cal N}\)) = kernal(\({\cal K}(T)\))

\({\cal K}(T) = \left\langle\left[\begin{array}{cc} -1 \\ 1 \\ 1\end{array}\right] \right\rangle\)

Since kernal \({\cal K}(T)\) is not trival, according to theorem KILT, Linear Transformation \(T\) is not injective.