Vector spaces

[mystic drift]

Will the matrix in 4(a) form a vector space?
Its not closed under scalar multiplication, right?

For 4(b), is the answer the standard basis for this matrix? And the dimension is the number of columns?

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[half temple]

4a: scalar multiplication isn't needed. Just existence of zero and closure under linear combinations.
4b: any basis will work, but the standard one is, of course, the simplest. And no, the dimension of a vector space is the number of basis vectors.

[mystic drift]

4a. Do you mean closure under addition like u+v=... where u,vεV
Or, like αu+βv=... where α,β are scalars

Is this even a vector space or no?

4b. So, if I just use the standard bases, they would be 6 in number. The dimension is 6 here then?

[mystic drift]

@half temple

[half temple]

4a.
What I mean is that if u and v are elements of a vector space, then so is au + bv, where a and b are arbitrary constants.
4b. Yes.

[mystic drift]

4a. If I do that, then the '1's in the symmetric matrix given become 2
That's violation of the condition right? Or can I simply divide it by 2 again to preserve the form like how we do during elementary row operations

[half temple]

Yes.
We can also show it like this:
{{a, 1}, {1, b}} = a{{1, 0}, {0, 0}} + b{{0, 0}, {0, 1}} + {{0, 1}, {1, 0}}
That last term is a problem, as it's a non-zero constant. If we didn't have it, we would have a linear space with basis vectors corresponding to the first two terms. With that term, however, it's a linear manifold, not a vector space.

[mystic drift]

Thank you so much!

[half temple]

You're welcome!

[mystic drift]

+close