Proving Vector <a,1,1> is a Subspace in R3

[Offlinedoctor]

For vector such as <a,1,1>, to prove its a subspace in r3, is it alright to immediately assume its not a subspace, as it doesn't meet the zero vector condition in that, 1=\=0?

And is there a specific way to set out subspace questions? I seem to just use intuition and two or three lines which worries me...

 

[Simon Bridge]

Did I read that right?
You want to prove that <a,1,1> is a subspace by assuming that it is not a subspace?
You mean it is not a vector space?

If <O,1,1> is the zero vector, then:
<O,1,1>+<a,1,1>=<a,1,1> => <O,1,1>=<0,0,0> = false, therefore <a,1,1> is not a vector space.

You have not "assumed" it is not a vector space, you have proven that it is not.

There is no standard approach for proving a subspace - just go through the definitions one at a time.
There are a lot of problems where there is no standard step-by-step procedure that will always produce the right answer. As you get better at math, your experience and understanding will replace and/or enhance your intuition.

 

[HallsofIvy]

Simon Bridge is correct. Saying " to prove its a subspace in r3, is it alright to immediately assume its not a subspace, as it doesn't meet the zero vector condition in that, 1=\=0?" makes no sense. If the problem is to prove it is NOT a subspace, then yes, saying there is no "a" such that <a,1, 1>= <0, 0, 0> so there is no zero vector in the set is sufficient to prove (not assume) that the set is not a subspace.