Linear algebra proof subspaces

In summary: And if you know XA=AX, isn't it not too hard to show cXA=cAX?In summary, the conversation discusses proving that the set W = {X : XA = AX} is a subspace of the vector space M2,2. The conversation mentions the closure conditions for a subspace and discusses the scenarios where AX=XA. The conversation also mentions that it is straightforward to show that cu is in W by multiplying the matrices. Finally, the conversation suggests that it is not difficult to show that (U+V)A=A(U+V) and cXA=cAX.
  • #1
oxlade15
2
0

Homework Statement


Let A be a fixed 2x2 matrix. Prove that the set W = {X : XA = AX} is a subspace of M2,2.


Homework 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.

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.
 
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  • #2
oxlade15 said:

Homework Statement


Let A be a fixed 2x2 matrix. Prove that the set W = {X : XA = AX} is a subspace of M2,2.

Homework 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.

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.

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)?
 

Related to Linear algebra proof subspaces

1. What is a subspace in linear algebra?

A subspace in linear algebra is a subset of a vector space that satisfies the following criteria: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication.

2. How do you prove that a set is a subspace?

To prove that a set is a subspace, you must show that it contains the zero vector, that it is closed under vector addition, and that it is closed under scalar multiplication. Additionally, you must show that the set is a subset of the original vector space.

3. What is the difference between a subspace and a vector space?

A subspace is a subset of a vector space that satisfies certain criteria, while a vector space is a set of vectors that are closed under vector addition and scalar multiplication. A vector space can contain subspaces, but a subspace cannot contain other subspaces.

4. Can a subspace be infinite?

Yes, a subspace can be infinite. As long as the subset satisfies the criteria of containing the zero vector and being closed under vector addition and scalar multiplication, it can be considered a subspace.

5. How do you prove that a subspace is a proper subset?

To prove that a subspace is a proper subset, you must show that it is a subset of the original vector space, but it is not equal to the original vector space. In other words, there must be at least one vector in the original vector space that is not contained in the subspace.

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