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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?

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2 Answers 2

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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 $5 - 4 = 1$.

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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.

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