# Finding a linear map…

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?

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

## 1 Answer

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.

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