I am given that the set $U = \{ (x_1, x_2, x_3, x_4, x_5 \in \mathbb{R}^5 : x_1 - 2x_2 = 0 \: \text{and} \: x_4 - 3x_5 = 0 \}$.
I have already shown that $U$ is a subspace of $\mathbb{R}^5$. Now I want to find a linear map whose null space is equal to $U$. My idea so far...
Define the linear map $T \in \mathcal{L}(\mathbb{R}^5)$ to be
$$T(x_1, x_2, x_3, x_4, x_5) = (x_1, 0, 0, x_4, 0).$$
Note that for $u \in U$ we have that $u = (2x_2, x_2, x_3, 3x_5, x_5)$. Hence, $Tu = 0$ and we conclude that null$\:T = U$.
Is this correct?
No, it is not correct. From $u=(2x_2,x_2,x_3,3x_5,x_5)$, what you can deduce is that $T(u)=(2x_2,0,0,3x_5,0)$, not that $T(u)=(0,0,0,0,0)$.
Consider$$\begin{array}{rccc}T\colon&\mathbb{R}^5&\longrightarrow&\mathbb{R}^2\\&(x_1,x_2,x_3,x_4,x_5)&\mapsto&(x_1-2x_2,x_4-3x_5)\end{array}$$instead.