# Finding the rank and nullity of transformation

I have this question:

Select each of the transformation below that is linear, has nullity 1 and rank 4.

A) $$T\begin{pmatrix} x \\y \\ z\\ t\\ \end{pmatrix}=\begin{pmatrix} x-t \\2z+3t\\t\\ \end{pmatrix}$$

B) $$T\begin{pmatrix} x \\y \\ z\\ t\\ \end{pmatrix}=\begin{pmatrix} x \\y\\z\\ \end{pmatrix}$$

C) $$T\begin{pmatrix} x \\y \\ z\\ t\\u\\ \end{pmatrix}=\begin{pmatrix} x \\y\\z\\u\\ \end{pmatrix}$$

D) $$T\begin{pmatrix} x \\y \\ z\\ t\\ u\\ \end{pmatrix}=\begin{pmatrix} x^2 \\y\\z\\u\\ \end{pmatrix}$$

My solution:

For A i got the rank as being 3

For B i got the rank as being 3

For C i got the rank as being 4 , so i have to find the nullity

For D its not linear

I am struggling to find the nullity of C. I know: the nullity is the dimension of kernel and i got the kernel as being x=0,y=0,z=0,u=0. However this would indicate a dimension of 0?

You're right that $$D$$ is not linear, so that can be ruled out. We can also rule out $$A$$ and $$B$$ because the codomain is $$\mathbb{R}^3$$, which means the rank is at most $$3$$. The answer is $$C$$ because clearly the range is all of $$\mathbb{R}^4$$, so the rank is $$4$$, and by the rank-nullity theorem, the nullity is $$5 - 4 = 1$$.
By the rank-nullity theorem, you know the nullity must be $$5-4 = 1$$.
The kernel consists of all elements with $$x = y = z = u = 0$$. The variable $$t$$ is free, so the kernel has dimension 1.