Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $(a_1, a_2, a_3) \mapsto (a_1, a_2, -a_1-a_2)$. I have to find $R(T), N(T)$ and a matrix that represents $T$. I know for my matrix that represents $T$ it will look something like this: $$ T= \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ -1 & -1 & 0 \end{pmatrix} $$ but I have no idea how to find the range and null space of $T$. I tried proving linearity and I couldn't even do that. The $-a_1 - a_2$ was throwing me off because it wasn't in terms of $a_3$ so I am not sure what to do.
2026-04-03 07:36:46.1775201806
Computing range, null space, and matrix of a linear transformation
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To find the null space literally set your ordered triple equal to zero and determine what are the valid values of $a_1, a_2, a_3$ need to be in order for the image to be (0,0,0).
Determining the null space of this linear transformation should tell you a lot about the range. Hint: What is required of the null space for the transformation to be injective?