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

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In summary: No, because that would mean that every element in H would also be in M. So we need to find another matrix that is in H but not in M. To do so, we can use the fact that A(B+C) = A(BC) (since B and C are in H). So we haveA(B+C) = A(BC)Since B and C are in H, this also means that (B+C) is also in H. So, (B+C) is also a matrix in H, so H is non-empty. Next, we need to show it is closed under addition. Since (B+C) is a matrix in H, this means
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student64
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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.
 
  • #3
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?
 

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

What is a subspace?

A subspace is a subset of a vector space that satisfies the three properties of closure under addition, closure under scalar multiplication, and contains the zero vector.

How do I determine if H is a subspace of Mnxn?

To determine if H is a subspace of Mnxn, you need to check if H satisfies the three properties of a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector. If all three properties are satisfied, then H is a subspace of Mnxn.

What is the zero vector?

The zero vector is a vector with all elements equal to zero. In the context of Mnxn, the zero vector is the nxn matrix with all elements equal to zero.

What is closure under addition?

Closure under addition means that when you add two vectors from a subspace, the result is also in the subspace. In the context of Mnxn, this means that when you add two nxn matrices from H, the result is also an nxn matrix in H.

What is closure under scalar multiplication?

Closure under scalar multiplication means that when you multiply a vector from a subspace by a scalar, the result is also in the subspace. In the context of Mnxn, this means that when you multiply an nxn matrix from H by a scalar, the result is also an nxn matrix in H.

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