How to prove that $A^{p/q}=(A^{1/q})^p$?

Question

defining matrix exponentiation for natural numbers by repeated multiplication and defining it for $\frac{1}{n}$ by: $A^{\frac{1}{n}}$ is the matrix s.t. $(A^{\frac{1}{n}})^n=A$. for a rational number $a=\frac{p}{q}$ $A^{a}=(A^{p})^{\frac{1}{q}}$ how do i prove that $A^a=(A^{\frac{1}{q}})^p$? I'm really unsure how to approach this, can i get some hints to the right direction?

Answer 1

Because it is problematic to even define exponentiation with non-natural exponents, you may want to restrict the general definition to matrices $A$ that are "close" to $I$, as in $(A-I)^n\to 0$ for $n\to \infty$. This allows us to define $$\ln A:=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(A-I)^n.$$ On the other hand, the series $$ \exp A:=\sum_{n=0}^\infty \frac1{n!}A^n$$ converges for all matrices $A$. Given these utilities, we can define $$ A^t:=\exp(t\ln A).$$ One can verify that $$ \exp(A+B)=\exp(A)\exp(B)\quad\text{if }AB=BA$$ and from this conclude that the expected rules hold.

Answer 2

Lets suppose all the matrices "behave nice" and everything is well defined (at least the matrices we are working with). By definition: $A^{\frac{p}{q}}:=(A^p)^{\frac{1}{q}}$. Denote $B:=A^{\frac{1}{q}}$. By definition, we know: $B^q=A \rightarrow (B^q)^p = B^{p\cdot q} = A^p$. So $B^p$ is a matrix with $(B^p)^q = A^p$ so again by definition: $B^p = (A^p)^{\frac{1}{q}}$.

So We get: $(A^{\frac{1}{q}})^p= B^p = (A^p)^{\frac{1}{q}}$

Q.E.D