# Finding the Null space and range of a transformation

I have the following transformation:

$$T: \mathbb{R}^3 \rightarrow \mathbb{R}^3\ \ \ \ T \begin{pmatrix} x\\ y\\ z\\ \end{pmatrix} = \begin{pmatrix} x+y+z\\ y+z\\ z \end{pmatrix}$$

I am trying to find the Null Space of $T$, $N(T)$ and the Range of $T$, $R(T)$.

I am still a little confused on the topic of transformation null space and range, but I do understand the concept of transformations and how they work.

My first guess is that the Null space would be something like: $$\begin{pmatrix} 0\\ x\\ x \end{pmatrix}, \begin{pmatrix} 0\\ x\\ y \end{pmatrix}$$

However, I'm not sure if this is correct, or if my understanding is a little flawed.

• If you know $T$ then the null space (or kernel) of the transformation is simply the solution set of the homogeneous system with coefficient matrix $T$. Mar 30 '16 at 16:51
• Oh, so it would be the solution to the Problem Ax=0, where $$A=\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}$$ ? Mar 30 '16 at 16:53
• That is correct. Mar 30 '16 at 16:54

Hint: the matrix of transformation (in canonical basis) is $M=\left(\matrix{1&1&1\\0&1&1\\0&0&1}\right)$ and $\det M=1$ ie, T is bijective.
You have the general form of the linear mapping. The basis of $\mathbb{R}^3$ is rather elementary (the standard basis). What you would like to do is write the transformation in terms of some matrix that can express it and find the solution to $Ax = 0$, that is the nullity of $T$.
One of the matrices you could express this transformation as is: $$A = \begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$$