Proving Subspace of Mnn: AB=BA for Fixed nxn Matrix B

  • Thread starter derryck1234
  • Start date
  • Tags
    Subspace
AB = BA then cA+dB=B(cA+dB)=cBA+dB=B(cA)+B(dB)=c(BA)+d(BB)=c(AB)+dB=AB(c+d)=ABc+ABd=BAc+BAd=A(c+d)B so it satisfies the condition too.In summary, the set of all n x n matrices A such that AB = BA for a fixed n x n matrix B is a subspace of Mnn because it satisfies closure under scalar multiplication and addition.
  • #1
derryck1234
56
0

Homework Statement



Prove that the set of all n x n matrices A such that AB = BA for a fixed n x n matrix B, is a subspace of Mnn.

Homework Equations



u + v is in the same vector space as u and v.
ku is in the same vector space as u, where k is any real number.

The Attempt at a Solution



I am drawn to think of a diagonal matrix when I think of this question. And if I multiply a diagonal matrix by a scalar, it can only be a diagonal matrix or the zero matrix, either way, it AB still equals BA. Similarly, adding two diagonal matrices obtains either another diagonal matrix or the zero matrix...so, in this way, A is a subspace of Mnn.

Am I correct?
 
Physics news on Phys.org
  • #2
i would just test the subspace requirements directly:

clearly the zero matrix is n the set
now say A,B satisfy AB = BA then is cA+dB in the set? that will satisfy closure under scalar multiplication and addition
 

FAQ: Proving Subspace of Mnn: AB=BA for Fixed nxn Matrix B

What is the definition of a subspace?

A subspace is a subset of a vector space that is closed under addition and scalar multiplication. This means that for any two vectors in the subspace, their sum and scalar multiples must also be in the subspace.

How do you prove that a set is a subspace?

To prove that a set is a subspace, you must show that it satisfies the three properties of a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector. In this specific case, you must also show that the set satisfies the additional property of commutativity, meaning that for any two matrices A and B, AB=BA.

Why is proving subspace of Mnn: AB=BA for fixed nxn matrix B important?

Proving subspace of Mnn: AB=BA for fixed nxn matrix B is important because it helps to determine whether the set of all matrices that commute with a fixed matrix B is a subspace. This has applications in linear algebra, where understanding subspaces is crucial in solving systems of linear equations and in studying vector spaces.

What are some examples of subspaces?

Some examples of subspaces include the set of all 2x2 matrices, the set of all polynomials of degree 2 or less, and the set of all real numbers greater than 0. In each of these examples, the set satisfies the properties of a subspace and is a subset of a larger vector space.

Can a subspace have infinitely many dimensions?

Yes, a subspace can have infinitely many dimensions. For example, the set of all polynomials of degree n or less is a subspace of the vector space of all polynomials, and it has infinitely many dimensions depending on the value of n. This is because there are infinitely many possible coefficients for each degree of the polynomial.

Similar threads

Vector Subspaces: Determining U as a Subspace of M4x4 Matrices
Replies
8
Views
1K
Can Direct Sums and Projections Fully Describe Subspaces in Linear Algebra?
Replies
5
Views
1K
Proving a Matrix to be nonsingular
Replies
26
Views
3K
MHB Proving Properties of 2x2 Matrices
Replies
5
Views
2K
A real parameter guaranteeing subspace invariance
Replies
26
Views
2K
Given specific v, dimension of subspace of L(V, W) where Tv=0?
Replies
15
Views
2K
Upper trianglar matrix is a subspace of mxn matrices
Replies
1
Views
2K
Linear algebra: Prove that the set is a subspace
Replies
8
Views
2K
T is cyclic iff there are finitely many T-invariant subspace
Replies
4
Views
2K
Prove that diagonal matrices are symmetric matrices
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top