Page 460, Exercise C25

Define the linear transformation

\(T: \mathbb{C}^3 \to \mathbb{C}^2, \hspace{2mm} T\left(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\right) = \begin{bmatrix}2x_1 - x_2 + 5x_3\\ -4x_1 + 2x_2 - 10x_3\end{bmatrix}\).

Find a basis for the kernel of \(T, \ \mathcal{K}(T)\). Is \(T\) injective?

Set \(T(x) = 0\), and then row reduce.

\[ \left[\begin{array}{ccc|c} 2 & -1 & 5 & 0 \\ -4 & 2 & -10 & 0 \end{array} \right] \]

library(pracma)

A <- matrix(c(2, -1, 5, 0,
              -4, 2, -10, 0),
             nrow = 2, ncol = 4, byrow = T)

a <- rref(A)
a
##      [,1] [,2] [,3] [,4]
## [1,]    1 -0.5  2.5    0
## [2,]    0  0.0  0.0    0

Which gives us the system of equations:

\[\begin{align*} x_1 - \frac{1}{2}x_2 + \frac{5}{2}x_3 &= 0 \\ 0 &= 0 \end{align*}\]

So \(x_1 = \frac{1}{2}x_2 - \frac{5}{2}x_3\), with \(x_2\) and \(x_3\) being free variables.

Thus, the basis set for the null space is \(\left\{\begin{bmatrix}\frac{1}{2}\\1\\0\end{bmatrix}, \begin{bmatrix}\frac{-5}{2}\\0\\1\end{bmatrix}\right\}\)

Theorem KILT says \(T\) is injective iff \(\mathcal{K}(T)=\{0\}\).

Since the kernel is not trivial, i.e. \(\mathcal{K}(T)\neq\{0\}, \ T\) is not injective.