Question

Let M₂ₓ₂ be the vector space of 2×2 matrices with real number entries.

Prove that if we fix any matrix A∈M₂ₓ₂ then the subset
C = {X∈M₂ₓ₂ ∣AX = XA}
is a subspace of M₂ₓ₂.

Answer 1

To prove that the subset C = {X∈M₂ₓ₂ ∣AX = XA} is a subspace of M₂ₓ₂, we need to show that C is closed under addition and scalar multiplication, and that it contains the zero vector.

Step-by-step explanation:

We need to prove that the subset C = {X∈M₂ₓ₂ ∣AX = XA} is a subspace of M₂ₓ₂. To prove this, we need to show that C is closed under addition and scalar multiplication, and that it contains the zero vector.

Let X and Y be two matrices in C, and let c be a scalar. We want to show that X + Y and cX are also in C.

First, we have (X+Y)A = XA + YA = AX + AY = A(X+Y), which shows that X+Y is in C.

Second, we have (cX)A = c(XA) = c(AX) = A(cX), which shows that cX is in C.

Finally, since the zero matrix O satisfies AO = O = OA for any matrix A, we can conclude that the zero matrix is in C.

Therefore, C is closed under addition and scalar multiplication, and it contains the zero vector, so it is a subspace of M₂ₓ₂.