# linear algebra subspace

could anyone help me on this subspace test please. would be much appreciated thank you

Let (V,+, ·) be $R^3$ with the usual vector addition and scalar multiplication. For each of the following, either use the subspace test to show that the given $W$ is a subspace of (V,+, ·) or explain why $W$ is not a subspace.

$$W := \{w \in V \mid w . (1,-3, 2) = 1\}$$

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## 2 Answers

You have a plane with normal vector $(1,-3,2)$ through the point $(0,0,1)$ that does not pass through the origin, and hence is not a subspace of $\Bbb{R}^3$.

Here is a plot of what the plane looks like.

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Firstly, (0,0,0).(1,-3,2)=0, not 1. Secondly, (1,-2,-3).(1,-3,2)=1 and (2,-3,-5).(1,-3,2)=1, but (3,-5,-8).(1,-3,2)=2. Lastly, (2,-4,-6).(1,-3,2)=2, so none of the 3 subspace conditions hold. Obviously, you need to argue with any one of the above, since a subspace requires all the 3 conditions simultaneously! However, if 1 were replaced by 0 in the question, the resulting subset would have been a subspace, the plane perpendicular to (1,-3,2).

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