# Linear algebra proof subspaces

1. Apr 28, 2012

1. The problem statement, all variables and given/known data
Let A be a fixed 2x2 matrix. Prove that the set W = {X : XA = AX} is a subspace of M2,2.

2. Relevant equations
Theorem: Test for a subspace
If W is a nonempty subset of a vector space V, then W is a subspace of V if and only if the following closure conditions hold.
1. If u and v are in W, then u + v is in W.
2. If u is in W and c is any scalar, then cu is in W.

3. The attempt at a solution
First off, we know that W is a nonempty subset of M2,2 since if you let X = the 2x2 zero matrix, then XA = AX.

I'm not certain about this, but I think that there are only 3 scenarios where AX=XA and that is if X is the identity, if X is the zero 2x2 matrix or if the matrix X = the matrix A.

Next, we need to show that if u and v are W, then u + v is also in W and that cu is in W.
I'm not sure how to do show that u + v is in W. For the cw is in W part, I think you can simply multiply the matrices out and show that they are equal, which is straightforward.

2. Apr 28, 2012

### Dick

For a fixed matrix A there aren't necessarily only those three scenarios. And if you know UA=AU and VA=AV, isn't it not too hard to show (U+V)A=A(U+V)?