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Suppose that $T:\mathbb{F}^3 \to \mathbb{F}$ is a linear transformation such that $$\text{null}(T)=\{(x_1,x_2,x_3)\in \mathbb{F}^3 | x_1=2x_2\} $$

Then prove that $T$ is surjective.

I saw this question in my book's exercise but I couldnt solve it. Can someone please give some hint about it. Thank you.

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2 Answers 2

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Actually $\text{null}(T)=\{(2x_2,x_2,x_3): x_2, x_3 \in \mathbb{F}\}=\{x_2(2,1,0)+x_3(0,0,1):x_2, x_3 \in \mathbb{F}\}$. In other words

$$\text{null}(T)=\text{span}\{(2,1,0), (0,0,1)\}$$

and $\{(2,1,0), (0,0,1)\}$ is linearly independent in $\mathbb{F^3}$. So you have $\dim(\text{null}(T))=2$, now you can apply the rank-nullity theorem to get $\dim(\text{rank}(T))=1$ making $T$ surjective since

$$\dim(\mathbb{F})=1=\dim(\text{rank}(T))$$

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Hint: Consider the fundamental theorem of linear transformations (i.e. the Rank-Nullity theorem). What is the dimension of $\text{Null}(T)$ and $\text{Rank}(T)$?

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  • $\begingroup$ I think dimension of $null(T)$ is 1 as the free variable is $x_2$ So the dimension of $rank(T)$ should be 2. I am not sure if this is correct $\endgroup$ Apr 3, 2021 at 9:22
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    $\begingroup$ @JalilAhmad $x_3$ is also free $\endgroup$
    – TheStudent
    Apr 3, 2021 at 10:03
  • $\begingroup$ @JalilAhmad Almost! $x_3$ is also a free variable as there is no restriction on it in the definition of $T$. So $\dim(\text{null}(T))=2$. $\endgroup$ Apr 3, 2021 at 10:09

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