Linear algebra proof subspaces

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SUMMARY

The discussion centers on proving that the set W = {X : XA = AX} is a subspace of the vector space M2,2, where A is a fixed 2x2 matrix. The proof utilizes the theorem for testing subspaces, which requires demonstrating closure under addition and scalar multiplication. It is established that W is nonempty since the zero matrix satisfies the condition XA = AX. The participants discuss the necessity of showing that for any matrices u and v in W, their sum u + v also belongs to W, and that for any scalar c, cu is in W.

PREREQUISITES
  • Understanding of vector spaces and subspaces
  • Familiarity with matrix multiplication and properties
  • Knowledge of linear algebra theorems, specifically the subspace test
  • Basic proficiency in working with 2x2 matrices
NEXT STEPS
  • Study the properties of matrix commutativity in linear algebra
  • Learn about the implications of the zero matrix in vector spaces
  • Explore examples of subspaces in M2,2 beyond the given scenario
  • Review scalar multiplication and its effects on matrix operations
USEFUL FOR

Students studying linear algebra, particularly those focusing on matrix theory and subspace properties, as well as educators teaching these concepts in a classroom setting.

oxlade15
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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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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)?
 

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