Uni Maths Questions

[Deleted User]

Thank you. That was what I was after

[Sach1_K]

I don't exactly know where to post this. Can you guys help me with it?
Thanks in advance
- Attached question

[Deleted User]

How do I find a basis of a subspace?

Eg. What is a basis for this: {(-1,2,0,4), (3,1,-1,2), (-5,3,1,6), (7,0,-2,0)} R^4

Thank you

[Alwin]

For the basis of a subset, put the vectors you have in matrix form:



Next, you find the row reduced echelon form, I'll assume that you know how to do this already:



So the column space is clearly the first two rows of the original martix

Thus, a basis is:



Hope that helps!

[Sach1_K]

I don't exactly know where to post this. Can you guys help me with it?
Thanks in advance
- Attached question

Don't worry, I got it now.

[Deleted User]

For the basis of a subset, put the vectors you have in matrix form:



Next, you find the row reduced echelon form, I'll assume that you know how to do this already:



So the column space is clearly the first two rows of the original martix

Thus, a basis is:



Hope that helps!

Thanks. Does this method work for all subspaces?

[Deleted User]

How do I show that the set (a,b,a-b,a+b) is closed under addition and scalar multiplication (R^4)?

[Alwin]

Thanks. Does this method work for all subspaces?

Yes, this called The Column Method of finding the basis for the span of a set of vectors given.

How do I show that the set (a,b,a-b,a+b) is closed under addition and scalar multiplication (R^4)?

Addition:
You can sub in values to test it:

which is in the set of since and

For a more general proof,
let and

by manipulation
Hence, the set is closed under addition

Scalar multiplication:
Again, you can sub in values to test it:

which is in the set of since and

For a more general proof,
let


Clearly the scalar, is also in the set

I think this proves it, but there may be a more elegant method. Refer to your textbook if you have one

[Deleted User]

How do I put the set of vectors {(2 1; 0 0), (0 0; 2 1), (3 -1; 0 0), (0 0; 3 1)} into a matrix with each vector as a column? Thanks

[Alwin]

How do I put the set of vectors {(2 1; 0 0), (0 0; 2 1), (3 -1; 0 0), (0 0; 3 1)} into a matrix with each vector as a column? Thanks

I'm not sure, because of your sparsely spaced semi-colons but is this what you mean?



If not, let me know

[Deleted User]

I meant that eg. (2 1; 0 0) =
2 1
0 0

[Deleted User]

Ok this is what I meant.

How do I write this set as a matrix? {, , , } (M_2,2.)

[mark_alec]

Consider the basis made up of {(1,0;0,0), (0,1;0,0), (0,0;1,0),(0,0;0,1)}.

[Deleted User]

I don't understand

[Yendall]

Q1: Which ordered pairs need to be added to the relation on the set to create the symmetric closure of ?

This is a really simple question, is this relation already symmetrically closed due to the fact that the symmetry of (a,a) is (a,a) ? Thanks.
Q2: Which ordered pairs need to be added to the relation on the set to create the equivalence relation generated by p?

Reflexivity: add and
Symmetry: add nothing?
Transitivity: Would I also add nothing?

This is where i'm confused. If the rule defines transitivity as , if there aren't any combinations of (x,z) in the set, do I need to add ordered pairs so that there are? If so, how many pairs is enough to make the entire set transitive?

Thanks for any help guize.