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?
Best Answer
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 $5 - 4 = 1$.