Linear Algebra: subspace etc?

Question

Why can't a subspace of a plane be any line, instead of just the lines passing through the origin? and related: Why can't a subspace of a space be any plane, instead of just the planes passing through the origin?

Answer 1

1) Because any subspace should contain a 0. For the line the 0 is the origin because (a,b) + (0,0) = (a,b).
2) The same reason: (0,0,0) is the only element with the property:
(a,b,c) + (0,0,0) = (a,b,c), for any pair (a,b,c)


Note: it is a nice exercise to show that the lines and the planes which pass through origin are really subspaces.

Answer 2

you ought to educate 3 issues: a million) The 0 vector is in the subset 2) The sum of two vectors X and Y in the subset should be contained in the subset 3) The scalar product ok*X, for any scalar ok, is contained in the subspace for any X contained in the subset 3 ordinarily follows promptly from 2, regardless of the indisputable fact that best in "organic" subspaces, so that you ought to to make it a dependancy of proving 3 and a couple of. on your impediment: a million) 0+0 = 0, so (0,0) is contained in the subset:) 2) if and are contained in the subset, then x1 + x2 = 0 and y1 + y2 = 0. consequently (x1 + x2) + (y1 + y2) = 0, (x1 + y1) + (x2 + y2) = 0, so is likewise in the subset. (you ought to also say that the subset is "closed less than addition") 3) if is in the subset, then x1 + x2 = 0, so ok*x1 + ok*x2 = 0, so ok* is contained in the subset consequently, the subset is properly a subspace! -- note: -- that is a quite rigorous data. i don't understand the position you're in the Linear Algebra route, yet ultimately, you'd be allowed to educate concerns like this far better swiftly and comfortably, utilising Linear Independence, foundation Vectors, Span of a Matrix, Kernel of the Homomorphism, etc.

Answer 3

To satisfy the axioms of a space, zero must be in the space.

For v in V a vector space over the reals.

then r*v in V, for any r in R
and
for any v1, v2 in V
v1+v2 in V

so, let v2=-1*v1
then
v1+v2 = v1+(-v1) = 0 is in V.

So zero must be in any vector space or subspace.

Answer 4

if u take any line y = mx +c
the points on the line will not satisfy the closure properties.