Transforming Matrix Addition to Multiplication without Exponentiation

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The discussion centers on the transformation of matrix addition into multiplication without using exponentiation. Participants clarify that while the equation e^{A+B} = e^A e^B holds true under the condition that matrices A and B commute (AB = BA), exponentiation leads to infinite results, unlike finite matrix multiplication. A proposed solution for expressing A.B in terms of exponentials is X = ABe^{-(A+B)}, although this approach is noted to be restrictive due to its lack of distributive properties.

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Suppose two matrices A and B . i want to transform matrix addition to matrix multiplication. e.g A+B into A.B.
Can anybody please tell me of any way i can do it ,except exponentiation? exponentiation gives an infinite answer.
 
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aliya said:
Suppose two matrices A and B . i want to transform matrix addition to matrix multiplication. e.g A+B into A.B.
Can anybody please tell me of any way i can do it ,except exponentiation? exponentiation gives an infinite answer.

It is certainly true that e^{A+ B}= e^Ae^B but what do you mean by "exponentiation gives an infinite answer"?
 
HallsofIvy said:
It is certainly true that e^{A+ B}= e^Ae^B but what do you mean by "exponentiation gives an infinite answer"?

e^{A+ B}= e^Ae^B if and only if AB = BA
 
trambolin said:
e^{A+ B}= e^Ae^B if and only if AB = BA

Don't A and B also need to be n x n matices?
 
Last edited:
Thankyou all for your replies.
"By exponentiation gives an infinite answer" i meant that suppose two matrices A and B, their multiplication A.B gives a finite answer i.e another finite matrix. but e^A.e^B give an infinite answer. doesn't it? Also can you please tell if there exists any X such that A.B=X(e^A.e^B)?
 
aliya said:
Thankyou all for your replies.
"By exponentiation gives an infinite answer" i meant that suppose two matrices A and B, their multiplication A.B gives a finite answer i.e another finite matrix. but e^A.e^B give an infinite answer. doesn't it?
No, it doesn't. If A and B are n by n matrices, then so are e^A, e^B, and e^Ae^B.

Also can you please tell if there exists any X such that A.B=X(e^A.e^B)?
 
ok thanks but " can you please tell if there exists any X such that A.B=X(e^A.e^B)?"
 
?? Of course their is:
X= ABe^{-(A+ B)}
 
I think he is trying to come up with a logarithm function. But that would be quite restrictive since the function wouldn't be distributive in general.
 

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