2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 $T$. $\endgroup$ Mar 30 '16 at 16:51
  • $\begingroup$ Oh, so it would be the solution to the Problem Ax=0, where $$A=\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}$$ ? $\endgroup$ Mar 30 '16 at 16:53
  • $\begingroup$ That is correct. $\endgroup$ Mar 30 '16 at 16:54
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Hint

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: $$A = \begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.