If two matrices $A$ and $B$ are commutative then all rules for real numbers $a$ and $b$ apply for the matrices?
For example, if $AB=BA$ then:
$(A+B)^2=A^2 + 2AB + B^2$
$A^3 - B^3 = (A-B)(A^2+AB+B^2)$
and so on...If the matrix $A$ is invertible then is $A^m A^n = A^{(m+n)}$, where $m,n$ are integers?
Commutative matrix
1
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linear-algebra
matrices
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4(1) The evaluation map $\Bbb C[x,y]\to \Bbb C[A,B]$ will be a ring homomorphism if $A,B$ commute. (I think this is the correct algebraic language for what you speak of.) (2) This holds very generally, not just for $A$ invertible (ie arbitrary monoids). – 2012-08-12
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0@anon: note that $m$ and/or $n$ might be negative. – 2012-08-12
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0[Oh, right; then in arbitrary groups.] – 2012-08-12
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0yes, you can show it by direct computation for the first identity. – 2012-08-12