Let \(T: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}\) be given by \(T(\begin{bmatrix} x \\ y \\ z \end{bmatrix}) = \begin{bmatrix} 2x + y + z \\ x - y + 2z \\ x + 2y - z \end{bmatrix}\). Find \(K(T)\). Is \(T\) injective?
To begin with, T should have its value be equivalent to zero so it can be set to row-reduced echelon form and then solved.
\(T(\begin{bmatrix} x \\ y \\ z \end{bmatrix}) = 0\)
Therefore…
\(\begin{bmatrix} 2x + y + z \\ x - y + 2z \\ x + 2y - z \end{bmatrix} = 0\)
Using the above, we can put this system of equations into
\(\begin{bmatrix} 2 & 1 & 1 \\ 1 & -1 & 2 \\ 1 & 2 & -1 \end{bmatrix}\)
We can put this into reduced row echelon form.
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 1 & -1 & 2 \\ 1 & 2 & -1 \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 1-(2\div2) & -1-(1\div2) & 2-(1\div2) \\ 1 & 2 & -1 \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 1-1 & -1-\frac{1}{2} & 2-\frac{1}{2} \\ 1-1 & 2-\frac{1}{2} & -1-\frac{1}{2} \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 0 & -1\frac{1}{2} & 1\frac{1}{2} \\ 0 & 1\frac{1}{2} & -1\frac{1}{2} \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 0 & \boxed{-1\frac{1}{2}} & 1\frac{1}{2} \\ 0 & 1\frac{1}{2} & -1\frac{1}{2} \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 0 & \boxed{-1\frac{1}{2}} & 1\frac{1}{2} \\ 0 & 1\frac{1}{2} & -1\frac{1}{2} \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 0 & \boxed{-1\frac{1}{2}} & 1\frac{1}{2} \\ 0 & 1\frac{1}{2}+-1\frac{1}{2} & -1\frac{1}{2}+1\frac{1}{2} \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 0 & \boxed{-1\frac{1}{2}} & 1\frac{1}{2} \\ 0 & 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 0 & \boxed{-1\frac{1}{2}}\div-1\frac{1}{2} & 1\frac{1}{2}\div-1\frac{1}{2} \\ 0 & 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1 & 1 \\ 0 & \boxed{1} & -1 \\ 0 & 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1-1 & 1--1 \\ 0 & \boxed{1} & -1 \\ 0 & 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2} & 1-1 & 1+1 \\ 0 & \boxed{1} & -1 \\ 0 & 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} \boxed{2}\div2 & 0 & 2\div2 \\ 0 & \boxed{1} & -1 \\ 0 & 0 & 0 \end{bmatrix}\)
This gives us the reduced row echelon form of the original system of equations as:
\(\begin{bmatrix} \boxed{1} & 0 & 1 \\ 0 & \boxed{1} & -1 \\ 0 & 0 & 0 \end{bmatrix}\)
Using the reduced row echelon form, we get the following system of equations…
\(x + 0y + z = 0\) and \(0x + y - z = 0\)
This can be reduced to
\(x + z = 0\) and \(y - z = 0\)
Which gives us the solutions
\(x = -z\) and \(y = z\)
This can be written as:
\(K(T) = \Bigg\langle\begin{Bmatrix}\begin{bmatrix}-1 \\ 1 \\ 1 \end{bmatrix}\end{Bmatrix}\Bigg\rangle\)
Since \(K(T) \neq \langle\begin{Bmatrix}\begin{bmatrix}0\end{bmatrix}\end{Bmatrix}\rangle\), \(K(T)\) is not trivial. As per the theorem KILT in the textbook, \(T\) is injective if and only if \(K(T)\) is trivial. \(K(T)\) is not trivial, therefore \(T\) is not injective.