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I am a University undergraduate student.

I have been asked to consider $A \subseteq \Re ^3$. Where $A = \{(x,y,z): x+y+z=1 \}$.

The question: Is $A$ a subspace of the Euclidean vector space $\Re ^3$, give reason for the answer.

I know that you need to determine whether $A$ satisfies the three properties of a subspace of $\Re ^3$.

However when looking at property 1 (zero vector of $\Re ^3 \in A$), I think that the zero vector isn't an element of $A$ as taking $x=y=z=0$, this doesn't satisfy $x+y+z=1$.

However I am cautious that I am wrong.

Any verification would be most grateful!

2 Answers 2

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Hint:

A subspace of a linear space always contains the zero vector...

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    So you are saying that this property is always satisfied as the question states that $A \subseteq \Re ^3$ – 2017-02-08
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    No, it is clearly not *always* satisfied as, for example, **your** $\;A\;$ does **not** fulfill it ... Read carefully and think. – 2017-02-08
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    Okay thanks for the help. I am still trying to get my head around this way of thinking. It can become very confusing if you don't break the problem down. Do you have anything I could read to point me in the correct direction when trying to tackle these sorts of questions? – 2017-02-08
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    @Ben There are millions (not exaggerating...) of links in linear algebra. Try, for example, the following one: http://linear.ups.edu/html/section-S.html – 2017-02-08
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Since $\vec0 \notin A \Rightarrow \langle A, + \rangle$ hasn't a neutral element and therefor cannot be a Group. Let alone an abelian Group for $\langle A, +,\cdot \rangle$ to be a vector space. Generally to check if a subset $U$of a vector space is a subspace. You only need to check if $\vec 0\in U$ and that $U$ itself is a Vectorspace over the same field.