Define the linear transformation \[T: C^{3} \rightarrow C^2\] \[T \left( \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[ \begin{array}{c} 2x_{1}-x{2}+5x_{3} \\ -4x_{1}+2x_{2}-10x_{3} \end{array}\right] \] The Kernel of a Linear Transformation Definition KLT says to suppose \(T: U \rightarrow V\) is a linear transformation.
Then the kernel of \(T\) is the set \(K(T)=\{u \in U | T(u)=0\}\)
Find vectors \(u\) such that \(T(u)=0\) \[ \left[ \begin{array}{c} 2x_{1}-x_{2}+5x_{3} \\ -4x_{1}+2x_{2}-10x_{3} \end{array}\right] = \left[ \begin{array}{c} 0 \\ 0 \end{array}\right] \] \[ = \left[ \begin{array}{ccc} 2 & -1 & 5 \\ -4 & 2 & -10 \end{array}\right] \] Row reduction \[ = \left[ \begin{array}{ccc} 2 & -1 & 5 \\ 0 & 0 & 0 \end{array}\right] \] \(2x_{1} -x_{2}+5_{3} = 0\)
\(x_{1} = \frac{1}{2}x_{2} - \frac{5}{2}x_{3}\)
We only have one pivot column with two free variables. To find a set of vectors to this system of equations we solve for \(x_{1}\) by giving \(x_{2}\) and \(x_{3}\) random values
\(x_{2} = 2\) and \(x_{3}=0\)
\(x_{1} = x_{2}\) \[ = \left[ \left(\begin{array}{c} 1 \\ 2 \\ 0\end{array}\right) \right] \]
Substituting different values \(x_{2} = 0\) and \(x_{3}=2\)
\[ = \left[ \left(\begin{array}{c} -5 \\ 0 \\ 2\end{array}\right) \right] \] Therefore,
\[K(T) = \left[ \left(\begin{array}{c} 1 \\ 2 \\ 0 \end{array}\right), \left(\begin{array}{c} -5 \\ 0 \\ 2 \end{array}\right) \right] \]
\(K(T) \neq 0\) so \(T\) is not injective.