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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?

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  • $\begingroup$ Hint: The null space of a matrix is the orthogonal complement of its row space. $\endgroup$ – amd Jan 1 '19 at 0:32
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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.

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  • $\begingroup$ Oh, I see now. Thank you very much! $\endgroup$ – Taylor McMillan Dec 31 '18 at 19:03

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