Matrices: Question about Commutativity

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The discussion centers on the commutativity of matrix operations involving scalars and quaternions. It confirms that scalars can be moved freely in matrix multiplication, allowing the expression X * kY to be simplified to k(X * Y). The conversation highlights the difference between commutative and non-commutative rings, specifically mentioning quaternions. The user is working with complex matrices and aims to demonstrate that Q * Q-1 equals the identity matrix, successfully applying the discussed principles. Ultimately, the manipulation of scalars in this context is affirmed as valid and effective.
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I have a question about commutativity.
I have two matrices X and Y and a constant k. I want to calculate X * kY. Can I bring k out the front to give k(X*Y)?
 
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Thanks for the link. It says "When the underlying ring is commutative, for example, the real or complex number field, the two multiplications are the same. However, if the ring is not commutative, such as the quaternions, they may be different."

Lol, I am actually working with quaternions. The matrices contain complex numbers. I am trying to show that
Q * Q-1 = Identity
but Q-1 is kX because its the inverse of a 2x2 matrix. So I thought it would be easier to work out QX then multiply the answer by k to (hopefully) give the Identity matrix (if that makes any sense at all).
 
Well then, I bow to people more knowledgeable than I about manipulating quarternions.
 
Same statement is true: a scalar (number), k, can be moved around pretty much as you wish. It is only multiplication of the matrices or quaternions that is non-commutative.
 
Thanks for the help, I moved k out the front and it worked :smile:
 

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