Finding the null space of given linear transformation $T: \mathbb{R}^{4}\mapsto \mathbb{R}^{3}$.

Question

In my Linear Algebra test, I was supposed to find the null space of the linear transformation $T:\mathbb{R}^{4}\mapsto \mathbb{R}^{3}$, which is defined as follows: $$T\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{bmatrix}=\begin{bmatrix}1 & 2 & 3 & 4 \\ 3 & 4 & 8 & 9 \\ 2 & 4 & 6 & 8\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{bmatrix}$$

Here is what I did. Since, null space is by definition the set $\{\textbf{x}\in \mathbb{R}^{4}:T(\textbf{x})=\textbf{0}\in \mathbb{R}^{3}\}$. So we just have to solve the homogeneous system $A\textbf{x}=\textbf{0}$. Since, the row $3$ is just the row $1$ rescaled by $2$ times, so we get the final coefficient matrix, after reducing the matrix into Echelon form as under:

$$\begin{bmatrix}1 & 0 & 2 & 1 \\ 0 & 2 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

So that we get $x_{1}=-2x_{3}-x_{4}$ and $2x_{2}+x_{3}+3x_{4}=0$. To my surprise, this set wasn't an option on the test. Could someone verify whether my solution is correct? Thanks in advance.

Answer 1

$A=\begin{bmatrix}1 & 2 & 3 & 4 \\ 3 & 4 & 8 & 9 \\ 2 & 4 & 6 & 8\end{bmatrix}$

$\mathrm{rref}(A)=\begin{bmatrix}1 & 0 & 2 & 1 \\ 0 & 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix}$

Clearly, $\dim(N(A)) =2$.

\begin{align}N(A)=\{c_1\begin{bmatrix}{-2}\\ {-}\frac{1}{2}\\ 1\\ 0\end{bmatrix}&+c_2\begin{bmatrix}{-1}\\ {-}\frac{3}{2}\\ 0\\ {1}\end{bmatrix} ,c_1,c_2\in \Bbb{R}\}\\\end{align}