# Linear algebra question on subspaces. Im terrible with these?

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Alright ive been struggling with these. THey re just very strange to me. Anyways theres a question here thats supposed to not be a subspace but i cant figure out why. Here are the questions

Determine whether the following sets form subspaces of R3

a) {(x1 x2 x3)T | x1 + x3 = 1}

b) {(x1 x2 x3)T | x1 = x2 = x3}

c) ({x1 x2 x3)T |x3 = x1 + x2}

d) {x1 x2 x3)T |x3 = x1 or x3 = x2}

so a and d are supposed to not be subspaces and b and c are. Im pretty sure i figured out the first 3 but i have no idea how to prove d isnt one. Im sure its a stupid easy answer but im just not seeing it. PLease help. (also if possible can you do the first 3 just so i know i did them correctly)

T is for transpose

Thanks!

• (I’ll just type row vectors with the understanding that they represent columns.)

a) Clearly this isn’t a subspace because (1, 0, 0) and (0, 0, 1) are in the set, but

(1, 0, 0) + (0, 0, 1) = (1, 0, 1)

is not.

b) These are just vectors in IR^3 of the form (a, a, a). A couple of quick computations show that if you add vectors of this form together, or multiply them by scalars, the form is unchanged.

c) These vectors have the general form (a, b, a+b). Again, you just have to show that adding such vectors together or multiplying them by scalars preserves this general form.

d) The problem here is the “or” in the condition. This set contains the vectors (1, 0, 1) and (0, 1, 1). But note that

(1, 0, 1) + (0, 1, 1) = (1, 1, 2).

The sum doesn’t satisfy the condition that x3 is equal to at least one of x1 or x2. So it is not closed under vector addition. This is one of the odd examples of a set closed under scalar multiplication but not under vector addition.

• The subspaces of R^3 are just the origin, lines through the origin, planes through the origin, and R^3 itself. Why? Any subspace has dimension 0 to 3, and any basis of these sizes generates these geometric structures. Perhaps more intuitively, start with between 0 and 3 vectors and see what happens when you allow arbitrary linear combinations of those vectors. The spaces you generate are subspaces, which are of the above 4 types.

In any case, for (d) to be a subspace it must be closed under arbitrary linear combinations. Take points (1,0,1) and (0,1,1) in that set. Add them–that’s a linear combination–to get (1,1,2). Their sum isn’t in the set, so the set isn’t a subspace. Geometrically, the set is the union of two different planes, so it can’t be a subspace.

• blahb31 has the best ****, yet examine this out: she/he purely made an extremely reasonable stupid typo in z(a,b,c) = (az,ab,ac) it of direction could be (za, zb, zc), basically a sprint mixing of the quite some scalars there… additionally another very gentle undertaking is that a subspace additionally must be non-empty… so as that would truly be extra as a third situation on blahb31’s 2. yet lots of the time you do not likely could think of approximately that.

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