# Prove that the following linear transformation is surjective

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.

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

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

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

• 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 Apr 3, 2021 at 9:22
• @JalilAhmad $x_3$ is also free Apr 3, 2021 at 10:03
• @JalilAhmad Almost! $x_3$ is also a free variable as there is no restriction on it in the definition of $T$. So $\dim(\text{null}(T))=2$. Apr 3, 2021 at 10:09