Question comes from “A First Course in Linear Algebra” by Robert A. Beezer.
Problem Description:
Let \(T:\mathbb{C}^3 \rightarrow \mathbb{C}^3\) be given by \(T\left(\begin{bmatrix}x \\ y \\ z\end{bmatrix}\right) = \begin{bmatrix}2x+y+z \\x-y+2z \\x+2y-z \end{bmatrix}\). Find \(\mathcal{K}(T)\). Is \(T\) injective?
Solution:
The kernel of \(T\) (\(\mathcal{K}(T)\)) is defined as the set of vectors \(\mathbf{x} \in \mathbb{C}^3\) such that \(T(x)\) is equal to the zero vector. In other words, we must find all cases in which:
\[ \begin{align} \begin{bmatrix}2x+y+z \\x-y+2z \\x+2y-z \end{bmatrix} &= 0 \\ \begin{bmatrix}2&1&1 \\1&-1&2 \\1&2&-1 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}&= \begin{bmatrix}0\\0\\0\end{bmatrix} \end{align} \]
To solve the above equation, we can first row reduce our linear transformation matrix:
\[ \begin{align} T &\Rightarrow \frac{1}{2}R_1 &= \begin{pmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ 1 & -1 & 2 \\ 1 & 2 & -1 \end{pmatrix} \\ T &\Rightarrow R_2 -R_1 &= \begin{pmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ 0 & -\frac{3}{2} & \frac{3}{2} \\ 1 & 2 & -1 \end{pmatrix} \\ T &\Rightarrow R_3 - R_1 &= \begin{pmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ 0 & -\frac{3}{2} & \frac{3}{2} \\ 0 & \frac{3}{2} & -\frac{3}{2} \end{pmatrix} \\ T &\Rightarrow -\frac{2}{3}R_2 &= \begin{pmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & -1 \\ 0 & \frac{3}{2} & -\frac{3}{2} \end{pmatrix} \\ T &\Rightarrow R_3 - \frac{3}{2}R_2 &= \begin{pmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \\ T &\Rightarrow R_1 - \frac{1}{2}R_2 &= \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \end{align} \]
Thus, based off our final reduced matrix our equation to solve updates to:
\[ \begin{equation} \begin{bmatrix}1&0&1 \\0&1&-1 \\0&0&0 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix} \end{equation} \]
providing us with:
\[ \begin{align} x + z = 0 \Rightarrow x &= -z \\ y - z = 0 \Rightarrow y &= z \\ z &= z \end{align} \]
Thus, we have that \(\mathcal{K}(T)\) is equal to the vector space in which \(y = z\) and \(x = -z\). Choosing \(z=1\) to get a basis vector, we have:
\[ \begin{equation} \mathcal{K}(T) = \text{Vect}\left\{\begin{pmatrix} -1\\1\\1\end{pmatrix}\right\} \end{equation} \]
Because the kernel of \(\mathcal{K}(T) \neq \emptyset\), we can also conclude that \(T\) is not an injective transformation.