Is there any formula for finding the power (integral as well as fractional) of the product of two matrices.
Equality (AB)^n=(B^n)(A^n) is incorrect in general. Corresponding example one can find for the case n=2.
The formula works for commuting matrices, i.e. if AB=BA. In the other case (AB)2=ABAB, and in general you cannot reorder the multipliers and get A2B2. As an example take
A whose first row is 0 0 and second row is 1 0, and
B whose first row is 0 1 and second row is 0 0.
Another example give matrices
A whose first row is 0 1 and second row is 1 0
B whose first row is 1 o and second row is 0 0
However, the formula works for commuting matrices, i.e. if AB=BA. In the fractional case, does it work ?
Dear P. Baliarsingh,
I think you should clarify what do you mean by fractional degree of matrix.
Dear Prof. Vladislav Babenko,
For the fraction "a" and two commuting matrices A,and B, is it "(AB)^a =B^aA^a" true. ? ( In particular, for a= 1/2 power is called the square root of a matrix and does it work ?)
Dear Professor Baliarsingh,
The theoretical background for your question is the theory of functions of self-adjoint operators (matrices). The fractional powers are correctly defined for Hermit-symmetric non-negative matrices, i.e. for matrices A that satisfy condition $ \ge 0$ for all vectors x. For such a matrix A and for q = n/m and function f(t) = t^q the matrix f(A) is the unique non-negative matrix G that satisfies equation $G^m = A^n$. If two matrices A, B commute, then any functions f(A) and g(B) of these matrices commute as well. From this you can prove the desired formula $A^qB^q = (AB)^q$ for q = n/m just by elevating $A^qB^q$ in power m and checking that the resulting matrix equals (AB)^n.
All the best, Vladimir
Dear P. Baliarsingh,
when I recommended to clarify in what sense is understood the fractional degree of matrix, I meant exactly what Vladimir Kadets says in his last answer.
Dear Harish Kumar Kotapally,
On my opinion the answer of Serpil Halici is correct for integer n. Proof is almost obvious. Please try to find mistake in your calculation.
Dear Harish Kumar Kotapally,
Let A, B be square commutative matrices of order m. Then for arbitrary integer n the equality (AB)^n=B^nA^n holds.
Proof. (AB)^n=ABAB... AB (n times). Since the matrices commutate we can move all Instances of the matrix B to the beginning of the last expression. We obtain
(AB)^n=BB...BAA...A=B^nA^n. The proof is ended.
Dear Prof. Vladislav Babenko:
"If A, B be square commutative matrices of order n then (AB)^n = B^n A^n is hold for arbitrary integer n." I think so,.
But if A, B noncommutative matrices, What are necessary and sufficient conditions in order to (AB)^n=B^n A^n holds?
For two commutative matrices A,B of order n(integer), (AB)^n = B^n A^n. Which concludes that " Matrix commutation implies (AB)^n = B^n A^n", but what is about its converse ?
In general, for A, B in Mn,
AB \ne (not equal to) BA.
Consider
(AB)n = (AB)(AB)(AB)...(AB) n-times
We can not move any consecutive terms in the right hand side. Thus
(AB)n \ne BnAn.
In particular case, if the two given matrices are commute, that is, AB = BA, we can move them, and we get
(AB)n = BnAn,
also, we have
(AB)n = AnBn.
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