Thread: Subspaces in R4

Hey,

The basic idea behind subspaces, to give you some intuition, is that anything you multiply or add together in the space stays in the space. That's pretty much all you need.

Let's get a general feeling for what these spaces would look like.

a) For this space, you know you want the things to look like (3, 3, 3, 3), (11441, 11441, 11441, 11441) -- anything whose components are equal. The important thing to imagine here is the geometry of what's going on. If you lower down to R3, you can imagine each vector having some direction and magnitude. To get to any point in R3 (that is, to have a basis for R3), you need three vectors that are linearly independent. Linear independence in this case means that all three vectors are going in completely different directions. So, for instance, if you used three pencils, and put two of your pencils flat on your desk and a third one sticking up, you would just have to make sure the two pencils lying flat weren't pointing in the same direction. If you put them perpendicular to one another, you have the familiar three axes (x, y, z).

So how do you determine the direction? Well, scaling certainly doesn't affect direction -- and this is true for any dimension. But you can see that (3, 3, 3, 3) = 3*(1, 1, 1, 1), for instance. This would lead us to believe our subspace is one-dimensional, and indeed it is. We have that all our vectors are just a multiple of (1, 1, 1, 1), so that any vector whose components are equal is a basis for this space.

Does this make sense?

OK, so I think that helped for most of them.
I got:
a) n*(1,1,1,1)
b) x1*(1,0,0,0) + x2*(0,1,0,0) + x3*(0,0,1,0) + x4*(0,0,0,1)
c) i have no idea how to do 4D cross products...I'm sure there's another way that I don't know
d) x1*(1,0) + x2*(0,1) and (-x3-x5, -x4, x3, x4, x5)

do those look right? (except c...can I have some more help on that one?)

for b) it is "some" nullspace, but you would then have to produce a linear map for which it is the nullspace OF.

you might get some insight by looking at the 2-dimensional and 3-dimension cases first:

in R^2, we would be talking about the set {(x,y): x+y = 0}. this clearly has the basis {(1,-1)} (among other possibilities: in fact {(a,-a)} would be a basis as long as a is non-zero).

in R^3, we are talking about {(x,y,z): x+y+z = 0}. these vectors all have the form (x,y,-x-y), and hopefully it is clear to you, that we are free to pick x and y howsoever we please, but then z is determined. so one basis would be {(1,0,-1), (0,1,-1)}.

now see if you can guess what this space might look like for R^4.

for c) if a vector v = (a1,a2,a3,a4) is perpendicular to (1,1,0,0) and (1,0,1,1) this means that (a1,a2,a3,a4).(1,1,0,0) = 0,

that is a1+a2 = 0, and also that (a1,a2,a3,a4).(1,0,1,1) = 0, so that a1+a3+a4 = 0.

so the 2nd coordinate is the negative of the 1st coordinate, and the 4th coordinate is the negative of the sum of the first and third coordinates:

(a1,a2,a3,a4) of the form (a1,-a1,a3,-a1-a4). how many variables do we have to specify to know this vector (and then, how many basis elements should we have)?

so for b) is it just (w, x, y, -(w+x+y))? i was expecting something a lot more complicated

and for c), if I put it into a 2x4 matrix = (a,b) and eliminate, I get 2 free variables and a general solution of ((b-x3-x4), (a-b+x3+x4), x3, x4). Is that what they're looking for?

also, were a and d right?

almost...you are looking for basis.

so you need to say what the vectors (w,x,y,-(w+x+y)) are linear combinations of.

well, we have three variables, so we need to specify 3 pieces of information. that's one way to think of a basis vector: a piece of (vector) information, that is independent of other vectors (other pieces of information) in the basis.

so what would a basis vector that only used "w's" look like? wouldn't it be (w,0,0,-w)? well, to pick a specific vector, we need to specify a value for "w".

note that we can write (w,0,0,-w) as w(1,0,0,-1).

so can you guess 3 linearly independent vectors (a1,a2,a3,a4), (b1,b2,b3,b4) (c1,c2,c3,c4) such that:

(w,x,y,-(w+x+y)) = w(a1,a2,a3,a4) + x(b1,b2,b3,b4) + z(c1,c2,c3,c4)?

if so, that last equation shows spanning, and linear independence (together with spanning) shows it is a basis.

do you know what a basis is?

so it would be w(1,0,0,-1)+x(0,1,0,-1)+y(0,0,1,-1) right?

that is a linear combination of 3 vectors.....what are the basis elements?

Originally Posted by Deveno

that is a linear combination of 3 vectors.....what are the basis elements?

Oh...I think I get it. So a basis is just a set of vectors, so basically you take them out of the linear combination. So it would be {(1,0,0,-1), (0,1,0,-1), (0,0,1,-1)}?

yes. a basis is well, the "basis" (in the ordinary english meaning of the word) of linear combinations.

now, although you have the right set, you still need to prove that it is indeed a basis (in the mathematical sense of the word).

this means showing it spans the subspace in question (you have already done this, in post #7, because

for any (w,x,y,-(w+x+y)) in our subspace, we have: (w,x,y,-(w+x+y)) = w(1,0,0,-1) + x(0,1,0,-1) + y(0,0,1,-1), so the basis is a spanning set),

AND...that it is linearly independent, that is, you must show that if:

a(1,0,0,-1) + b(0,1,0,-1) + c(0,0,1,-1) = (0,0,0,0), that a = b = c = 0. this shouldn't be too hard.

for (c) you can take a similar approach. (d) is a bit more intense, because we are talking about some rather general types of subspaces.

for (a) you want to list the basis element, not a "typical" element of the subspace (although they are closely related since the subspace is 1-dimensional).