Explain why linear map is surjective

Question

Suppose T is a linear map from $ℝ^4$ to $ℝ^2$ such that the kernel of $T$ is $$\left\{\left.\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}\right|\quad x_1=5x_2\quad\text{ and }\quad x_3=7x_4\right\}$$

I have to explain why $T$ must be surjective. I know how to tell this from a rref matrix because the rank has to be the same as the dimensions of $ℝ^2$, which would be $2$. But I do not know where to start in this case.

Answer 1

Nullity$(T)=2=$Dimension of kernel of $T$ as we can take $x_2$ and $x_4$ as free. So by rank nullity theorem , $4=$rank$(T)+$Nullity$(T)$ which gives rank$(T)=2$ and hence surjective.

Answer 2

Since we know the kernel, it should be easy to find a basis, and thus the dimension of the null space. Then since $\dim \mathbb R^4 = 4$, we can simply apply the rank-nullity theorem to obtain the rank of $T$. If the rank of $T$ is indeed $2$, then we're done.