We know that the product $A^nA^m = A^{n+m}$, $A$ is a square matrix and $n,m$ are integers.
Is this correct for non-integer $A^{\frac{1}{n}} A^{\frac{1}{m}} = A^{\frac{1}{n}+\frac{1}{m}}$.
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Is the matrix diagonalizable? – Michael McGovern Jan 10 '17 at 22:25
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So $A^{\frac{1}{2}} A^{\frac{1}{2}}$ is not equal to $A$ pjs36. – B.Mohammed Jan 10 '17 at 22:31
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1That's not exactly what I meant. If you can diagonalize a matrix, it is computationally easier to raise it to a power--weather it is an integer or a rational number. – Michael McGovern Jan 10 '17 at 22:35
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2How are you defining fractional powers of your matrix? There may be many, possibly even infinitely many, matrices $B$ such that $B^n = A$. Which one do you choose to be $A^{1/n}$? – Robert Israel Jan 10 '17 at 22:47
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Michael McGovern, Is it computationally easier to raise it for a diagonal matrix (it seems that it easier to calculate it) or for all diagonalizable matrices. – B.Mohammed Jan 10 '17 at 23:07
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We don't even have $A^{1/2}A^{1/2}=A$. For instance, let $A$ be the $3\times 3$ identity matrix. Every rotation matrix $R$ through an angle $\pi$ satisfies $R^2=A$, so in a sense each is a square root of $A$. But the product of any two such distinct matrices is not $A$. – TonyK Jan 10 '17 at 23:37