How do we do 8b? Finding a subspace

[brazen wedge]

I'm stuck because after factoring x and y, to find the general solution fo W1, we need another vector in W3 to span 3D. so we need another basis in W3, but if it is a basis, W1 and W2's direct sum is the field because their intersection is the zero vector

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[brazen wedge]

Here's my working so far

[wintry cargo]

In theory you could take W2=F^3

But if you really want the subspace to be strict you could just take W2 a plane spanned by a vector of W1 and a vector that is not in W1

[strong steeple]

the first one's the easy one

[brazen wedge]

oh really?

[strong steeple]

find a basis for the subspace (x,y,x+y) and then extend it to a basis of F3

that will also give you a way to find W3

[brazen wedge]

oh so its linearly dependent? but that means there's no new basis so the dimension is still in F^2 no?

[strong steeple]

what is linearly independent?

[brazen wedge]

For a basis, it requires linearly independent and spanning. So basically independence means no vector can be written as a multiple of another (so no vector is redundant).

[strong steeple]

what is "it" in coffey's comment?

[strong steeple]

this one

[brazen wedge]

i dont see 'it'

[strong steeple]

.

yes you do

[brazen wedge]

oh my comment

[strong steeple]

regardless, name a basis for W1

[brazen wedge]

here's my working, the basis is there

when i say 'it' i meant the new basis we are adding for W_2

[strong steeple]

checks out

now extend this basis to a basis of F3

[brazen wedge]

how do we do that?

[strong steeple]

add a third vector such that the determinant is nonzero

any vector will do

[brazen wedge]

sorry i didnt learn determinants yet

could u give an example vector?

[strong steeple]

(1,0,0) for example

[wintry cargo]

a bit of a technicality but the map (x,y,z)—>(x,y,x+y) is a projection so it’s kernel and image (W1) are in direct sum but just extending is enough so it may be overkill

[strong steeple]

you can verify the resulting system is linearly independent

[brazen wedge]

oh yes, but then, wouldnt the direct sum of W1 AND W2 be only the 0 vector?

[strong steeple]

what the..?

what's union got to do with anything here?

[brazen wedge]

maybe i'm confused but they're saying the direct sum of W1 and W2 cant be the field^#?

[brazen wedge]

sorry i mean conjunction

[strong steeple]

intersection

[brazen wedge]

oh yes

[strong steeple]

this is part a

you take W2 = F3 and you're done

the sum W1 + W2 is not direct

[brazen wedge]

oh okay is that the only solution?

because honestly i wouldnt have come up with that

[strong steeple]

there are infinitely many ways to pick a basis

so, no, not the only solution

but it is A solution

[wintry cargo]

No

I was just saying it’s a possibility

[brazen wedge]

what other solutions are there tho?

[strong steeple]

infinitely many....

you're not gonna list all of them

[brazen wedge]

okay but i mean 1 eg

[strong steeple]

im off to work, hf coffey

[brazen wedge]

thanks for you help

[wintry cargo]

Like aL said you just need to complete your family of two vectors spanning W1 in a basis of F^3

[brazen wedge]

okay i'm a bit tired so i'll reread your messages for that part

for the next part, how would you find subspaces as well?

[wintry cargo]

If v_1 and v_2 are the two vectors spanning W_1, take a third vector v_3 that is independent of (v_1,v_2) such that (v_1,v_2,v_3) is a basis of F^3

Then set W_3=span(v_3)

[brazen wedge]

ohh okay i get that one aL said that's the hard one tho? is there a reason why?

[wintry cargo]

For The first one you don’t need the sum to be direct to you could take W_2=F^3

[brazen wedge]

ohh okay thanks for your help too!