Question

Linear algebra: subspaces?

Is the given subset of R^3 a subspace?
The set of all vectors of the form [a; b; c], where a+2b-c=0 is a subspace.

Answer: Yes

I don't get why it is a subspace because it satisfies the closure properties? Can someone please help me?

Answer 1

it doesn't satisfy**

Answer 2

\[\vec v=a+2b-c=0\]\[\vec u=x+2y-z=0\]add them together\[\vec v+\vec u=a+2b-c+(x+2y-z)=0\]\[\implies (a+x)+2(b+y)-(c+z)=0\]

Answer 3

\[\vec w=(a+x)+2(b+y)-(c+z)=0\]let\[\begin{array}aa+x=p\\b+y=q\\c+z=r\end{array}\]then\[\vec w=p+2q-r=0\in W\]therefor closed under addition

Answer 4

are you ok with that?
the scalar part is trivial

Answer 5

Yes, I understand it now. Thanks!

Answer 6

welcome :)