Determine if subset is subspace of R3. Need Help.

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Homework Help Overview

The discussion revolves around determining whether a given subset of R3, defined by the equation z = 3x + 2, qualifies as a subspace. Participants are exploring the necessary conditions for a subset to be considered a subspace, particularly focusing on the inclusion of the zero vector and the closure properties under addition and scalar multiplication.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning whether the zero vector is included in the subset and discussing the implications of this for the subspace criteria. There are inquiries about how to apply closure under addition and scalar multiplication, with some confusion regarding the correct approach to demonstrate these properties.

Discussion Status

The discussion is active, with participants providing feedback on each other's understanding of the subspace criteria. Some guidance has been offered regarding the need to consider specific elements of the set rather than arbitrary vectors from R3. However, there is no explicit consensus on the overall conclusion regarding the subset's status as a subspace.

Contextual Notes

Participants are grappling with the implications of the subset not containing the zero vector and how this affects the closure properties required for a subspace. There is an emphasis on the need to clarify the definitions and properties of the vectors within the context of R3.

digitol87
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1. {[x,y,z] | x,y,z in R, z = 3x+2}.
How do I determine if this subset is a subspace of R3? Am I wrong when I say this set contains the zero vector? If this is the case, then I have to use the addition and multiplication closure methods, right?

Thanks
 
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digitol87 said:
{[x,y,z] | x,y,z in R, z = 3x+2}.
How do I determine if this subset is a subspace of R3? Am I wrong when I say this set contains the zero vector?
Yes, wrong. (0, 0, 0) does not satisfy z = 3x + 2.
digitol87 said:
If this is the case, then I have to use the addition and multiplication closure methods, right?

Thanks
 
Mark44 said:
Yes, wrong. (0, 0, 0) does not satisfy z = 3x + 2.

OK, That's a good first step. I'm still a little confused as to how to apply the addition method.
can I add [x,y,z] + [a,b,c] then get [x+a, y+b, z+c] ? Then what?

Thanks.
 
No, you have to take two elements in the set, and show that their sum is in the set. You can't take any old arbitrary vector in R3. Also, it must be true that if v is an element in the set, then kv is also in the set.

What must be true for any element in your set? I.e., how do you distinguish between vectors in your set and plain old vectors in R3?
 

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