by rhys
Last Updated June 21, 2018 17:20 PM

Let U be the subset of $\mathbb{R}^3$ defined by:

$$U=(x,y,z):x=3y,y+z=1$$

How can I show $U$ is not a subspace of $\mathbb{R}^3$?

Can anyone help me with this problem?

**HINT**

Note that for $(x,y,z)=(0,0,0)$

- $x=3y \implies 0=0$
- $y+z=1 \implies 0=1$

Every subspace should contain the zero vector, and your subspace does not since $y + z = 0 + 0 = 0 \neq 1$.

When a problem asks you to show that "A" is or is not "X", look at the definition of "X"! A non-empty subset, A, of vector space V is a "subspace" if and only if A is a subset of V that is closed under vector addition and scalar multiplication. In particular, if v is in A then "closed under scalar multiplication" means -v is also in A. And "closed under vector addition" then means that v+ (-v)= 0 is in A. That is, as others have said, sufficient to show that this subset is not a subspace. (0, 0, 0) satisfies "x= 3y" but **not** "y+ z= 1".

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