Is Q a Subspace of R2 or R3?

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The discussion focuses on determining whether the set Q is a subspace of R2 or R3 under different definitions of Q. The first definition, Q = {(v1, v2, 0) | v1, v2 ∈ R}, meets the criteria for a subspace as it contains the zero vector and is closed under addition and scalar multiplication. The second definition, Q = {(v1, 0, 0) | v1 ∈ R}, also qualifies as a subspace for similar reasons. However, the third definition, Q = {(v1, v2) ∈ R2 | v1 = v2 + 1}, does not satisfy the subspace criteria since it does not include the zero vector. Overall, the discussion clarifies the application of subspace conditions to various cases of Q.
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Hey, this isn't for homework per se, but if anyone could lend me a hand figuring this out I'd appreciate it a lot!

Homework Statement



Determine whether Q is a subspace of R2/R3 in the following cases:

Homework Equations



Q = \{\left v = \left( v1, v2, 0 \right) | v1,v2 \in R \right\}

Q = \{\left v = \left( v1, 0, 0 \right) | v1 \in R \right\}

Q = \{\left v = \left( v1, v2 \right) \in R2 | v1 = v2 + 1, v1,v2 \in R \right\}

The Attempt at a Solution



I honestly am not sure where I'm meant to start here. I know there are 3 conditions where a subspace may be valid; when it's equal to zero, If X and Y are in U, then X+Y is also in U
and If X is in U then aX is in U for every real number a.

How exactly am I meant to go about applying those rules? If someone could set me off on the right track that'd be awesome.
 
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\{\left v = \left( v1, v2, 0 \right) | v1,v2 \in R \right\}

(0,0,0) is clearly in Q

Let x = (x1,x2,0) and y =(y1,y2,0) be vectors in Q

then x+y = (x1+y1, x2+y2, 0) is also in Q

and ax = a(x1,x2, 0) = (ax1, ax2, 0) is in Q
 
Tweet said:
Hey, this isn't for homework per se, but if anyone could lend me a hand figuring this out I'd appreciate it a lot!

Homework Statement



Determine whether Q is a subspace of R2/R3 in the following cases:

Homework Equations



Q = \{\left v = \left( v1, v2, 0 \right) | v1,v2 \in R \right\}

Q = \{\left v = \left( v1, 0, 0 \right) | v1 \in R \right\}

Q = \{\left v = \left( v1, v2 \right) \in R2 | v1 = v2 + 1, v1,v2 \in R \right\}

The Attempt at a Solution



I honestly am not sure where I'm meant to start here. I know there are 3 conditions where a subspace may be valid; when it's equal to zero, If X and Y are in U, then X+Y is also in U
and If X is in U then aX is in U for every real number a.
To correct your terminology, there are 3 conditions for verifying that a subset U of a vector space V is a subspace of that vector space. 1) Zero is an element of U. The other two are fine.
Tweet said:
How exactly am I meant to go about applying those rules? If someone could set me off on the right track that'd be awesome.

See RandomVariable's reply.
 
Cool, thanks very much guys. Makes sense now!
 

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