# Thread: prove a vector subspace

1. ## prove a vector subspace

I know i first need to prove that it is empty,

how do i show that the zero vector is in U?

i know I have to take two elements of U and add them and see if I still am in the same space, but really not sure how to?

any help appreciated.

2. ## Re: prove a vector subspace

To show the 0-matrix is in U, you need to verify that:

A0B = B0A, can you do this?

For your next question, suppose that:

AXB = BXA and AYB = BYA.

You must show that A(X+Y)B = B(X+Y)A.

Start by expanding the LHS to get:

A(X+Y)B = (AX + AY)B = AXB + AYB. Do the same for the RHS. What can you say about: A(X+Y)B - B(X+Y)A ?

3. ## Re: prove a vector subspace

Originally Posted by Deveno
To show the 0-matrix is in U, you need to verify that:

A0B = B0A, can you do this?

For your next question, suppose that:

AXB = BXA and AYB = BYA.

You must show that A(X+Y)B = B(X+Y)A.

Start by expanding the LHS to get:

A(X+Y)B = (AX + AY)B = AXB + AYB. Do the same for the RHS. What can you say about: A(X+Y)B - B(X+Y)A ?
so

AXB + AYB = BXA + BYA

AB( X+Y) = BA(X+Y) ?

does this prove it?

4. ## Re: prove a vector subspace

You don't prove something by starting out with what you're trying to prove. You start with what you're GIVEN.

We want to prove that if X and Y are in U, then X+Y is in U. Maybe this isn't true, we need to find out. What is our criterion for deciding if X,Y are in U? That AXB = BXA, and AYB = BYA. So that is where we start.

So the question is: is it THEN true that:

A(X+Y)B = B(X+Y)A?

Now A(X+Y)B = (A(X+Y))(B) (associative law for matrix multiplication)

= (AX+AY)B (distributive law for matrix muultiplication)

= AXB + AYB (distributive law again).

Since AXB = BXA (given),

= BXA + AYB ....can you continue?

It is NOT true that:

AXB + AYB = BXA + BYA implies AB(X+Y) = BA(X+Y), in general, you can't "treat matrices like numbers" and "change the order of multiplication" (bring stuff "in front"). AB and BA are usually two different matrices.

5. ## Re: prove a vector subspace

And if X in U, show kX in U, k a scalar.

6. ## Re: prove a vector subspace

Originally Posted by Hartlw
And if X in U, show kX in U, k a scalar.
Towards this end, it might be helpful to recall that:

kX = (kI)X and that the matrices kI DO commute with any matrix, so k(AXB) = (kI)(AXB) = A(kI)(XB) = A((kI)(X))B = A(kX)B.