Just check if it satisfies scalar multiplication and addition properties. Also check if the Zero vector is contained in your set(all though this is redundant.) Can you do this?
Is this a subspace, if so, why and how?
1) set of all vectors from R^n which have a first coordinate of 0.
2) set of all vectors from R^2 of form
with x - y =2
3) set of all vectors from R^3 of form
4) set of all vectors from R^5 satisfying x1 + 2x2 = x3 & x4=2x5
For instance, the first example is very simple:
Let (for easier notations) be an integer and the set of vectors in which have a first coordinate of 0.
Clearly, belongs to .
Let and be two elements in A. That means .
Let be a real number, then
Thus every linear combination of vectors in is in .
and is a subspace of
Only because just showing "if v in S, then av in S" is not enough to show that S is a subspace. You also have to show "if u and v in S then u+ v is in S".
Let u= and let v= [tex]\begin{bmatrix}y_1\\y_2\\y_1 + 2y_2\\2y_5\\y_5 \end{bmatrix}[/quote]
Then u+ v=
which is again of the same form.
By the way, Scopur, showing that the 0 vector is in the set is not entirely redundant. If S= {}, the empty set, then it is trivial that "if v is in S, then cv is in S" and "if u and v are in S then u+ v is in S" but the standard definition of subspace requires that it be non-empty. You must show that there is some vector in the set and it is usually easiest to show that the 0 vector is in the set.