Determine if H is a Subspace of Mnxn: Let A be an nxn Matrix

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SUMMARY

H, defined as H = {B ∈ Mnxn | AB = BA} for an nxn matrix A, is indeed a subspace of Mnxn. To establish this, one must demonstrate that H is non-empty, closed under addition, and closed under scalar multiplication. The identity matrix serves as a non-empty element of H, and the closure under addition is proven by showing that for any two matrices B1 and B2 in H, the sum (B1 + B2) also satisfies the condition AB = BA. The closure under scalar multiplication follows similarly.

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Let A be an nxn matrix and let H= {B E Mnxn|AB=BA}. Determine if H is a subspace of Mnxn.

This was a test question that I got incorrect. I didn't like the way my teacher proved this afterwards, they said it IS a subspace of Mnxn. Any help in explaining how it could be would be greatly appreciated.
 
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First, you need to show that H is contained within M (obvious; no work needed but for rigor, you should at least state that H is contained within M). I am assuming here that M, the space of nxn matrices, includes the operations + and * (addition and multiplication). You will need to show that H is closed under addition (i.e. any two matrices in H added together should result in a matrix that is also in H) as well as multiplication. I'll show you one step to give you an idea of what's needed.

Suppose B1 and B2 are matrices in H. We wish to show that (B1+B2) is also a matrix in H. To do so, we must verify that A(B1+B2) = (B1+B2)A. Let's start with the left hand side:

A(B1+B2) = A B1 + A B2 = B1 A + B2 A

(here we can replace A B1 with B1 A because B1 is an element of H, and hence satisfies AB1 = B1A; similarly for A B2).

Finally,

B1 A + B2 A = (B1+B2) A.

Hence, (B1+B2) satisfies the condition A(B1+B2) = (B1+B2)A. Therefore, (B1+B2) is also in H, so H is closed under addition. You finish the rest by showing H is closed under multiplication.
 
rs1n said:
First, you need to show that H is contained within M (obvious; no work needed but for rigor, you should at least state that H is contained within M).

H is constructed from elements of M, so clearly this is not necessary at all.

The only necessity to showing something is subspace is to show that it is non-empty, closed under addition and scalar multiplication.

What's the easiest way to show something is non-empty? Well, is the identity matrix in there?
 

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