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### Course: Linear algebra>Unit 1

Lesson 4: Subspaces and the basis for a subspace

# Linear subspaces

Introduction to linear subspaces of Rn. Created by Sal Khan.

## Want to join the conversation?

• Why do we define linear subspaces? What are they used for? And why are they closed under addition and scalar multiplication specifically (as opposed to only being closed under addition, for example)?
• There are I believe twelve axioms or so of a 'field'; but in the case of a vectorial subspace ("linear subspace", as referred to here), these three axioms (closure for addition, scalar multiplication and containing the zero vector) all the other axioms derive from it.
• What is the difference between subset and subspace?
• A subset is a term from set theory. If B is a subset of a set C then every member of B is also a member of C. The elements (members) of these sets may not be vectors, or even of the same type! For instance, set C could contain a blue teapot and a small horse.

A subspace is a term from linear algebra. Members of a subspace are all vectors, and they all have the same dimensions. For instance, a subspace of R^3 could be a plane which would be defined by two independent 3D vectors. These vectors need to follow certain rules. In essence, a combination of the vectors from the subspace must be in the subspace too.
• what does closure mean exactly?
• Closure means belongs to the same set. For instance, consider the set of integers. They are closed under addition. Adding an integer to another integer gives you an integer. Adding a vector in a subspace to another vector in a subspace gives you a vector which is also in the subspace, so subspaces are closed under addition too.

For an example of something which isn't closed, consider the division of integers. This is not always closed. In the case of 4 divided by 2, the result is 2. That's still an integer, but 4 divided by 5 is 4/5 which is not an integer. It is a rational number which doesn't belong to the set of integers. I'm rambling a bit, but hope it helps!
• Is a subspace of R2 always either zero-vector, or a line, or all of R2? Is there any other possible way to divide R2 that creates a subspace?

Is the overall point then that a point is a subspace of a line, a line is a subspace of a plane, a plane is a subspace of a 3-d space, a 3-d space is a subspace of an R4 "space", and so on, and there are no other valid subspaces?
• Technically speaking, R^2 is a subspace unto itself!

This is refereed to as an "improper subspace".
• Sal is claiming that Span(v_1, v_2, v_3) will always be a subspace of R^3. But what if the vectors are linearly dependant in which case Span(v_1, v_2, v_3) = R^2? R^2 isn't a subspace of R^3 is it? What's going on here?
• If the vectors are linearly dependent (and live in R^3), then span(v1, v2, v3) = a 2D, 1D, or 0D subspace of R^3. Note that R^2 is not a subspace of R^3. R^2 is the set of all vectors with exactly 2 real number entries. R^3 is the set of all vectors with exactly 3 real number entries. Saying R^2 is a subspace of R^3 is false because that's like saying a vector with exactly 2 entries also has exactly 3 entries (that's obviously wrong because 2 doesn't equal 3).
• Do you guys know of any webpage where I can do linear algebra exercises?
• Does it mean that all subspaces must pass origin? For exemple in R3 a plane must go trough origin otherwise it is not a valid subspace?
• I think so. Otherwise it does not contain the zero vector.
• In R2, does it mean that there are only 3 possible "forms"/"shape" for a span: a point (the set containing the zero vector only), a line, and a plane (the whole R2 itself) ?
• Yes, exactly. This is because the shape of the span depends on the number of linearly independent vectors in the set. The span of the empty set is the zero vector, the span of a set of one (non-zero) vector is a line containing the zero vector, and the span of a set of 2 LI vectors is a plane (in the case of R2 it's all of R2).
• So in R2, you can't have a subspace defined as,
V = (0,2) + t(1,2)
but you can have
V = t(1,2)
as a subspace of R2?
• Yes, that's correct, since your first V isn't closed under addition and multiplication.

Say we have V(t) = [0,2] + t[1,2]
If V is a subspace, the following must be true: V(a+b) = V(a) + V(b)
V(a+b) = [0,2] + (a+b)[1,2]
V(a) + V(b)
= [0,2] + a[1,2] + [0,2] + b[1,2]
= [0,2] + (a+b)[1,2] + [0,2]
= V(a+b) + [0,2]

So V(a+b) =/= V(a) + V(b) and V isn't a subspace.