# Proving the logarithm rule for the matrix exponential

I am currently doing some work out of Undergraduate Analysis by Serge Lang and I have run into a bit of trouble with this question on the matrix logarithm.

I showed that on the space of $$n\times n$$ real matrices, if $$|I-C|<1$$ we can define the matrix logarithm as follows: $$\log(C)=\sum (-1)^{n+1}\frac{(C-I)^n}{n}$$ As well as the standard power series definition of the matrix exponential.

The author then goes on to say "Show that if $$A,B$$ commute, then $$\exp(A+B)=\exp(A)\exp(B)$$ State and prove the similar theorem for the log." Proving the identity for the exponent was relatively easy, only requiring some use of the binomial theorem and some rearranging, however, I am not sure how to proceed with proving the identity for the matrix logarithm. I believe we want, if $$A$$ and $$B$$ commute,we get $$\log(AB)=\log(A)+\log(B)$$ We then get $$\log(AB)=\sum (-1)^{n+1} \frac{(AB-I)^n}{n}=\sum_{n} \frac{(-1)^{n+1}}{n}\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}A^kB^k$$ but I am unsure how to proceed from here. Anyone have any suggestions? So far we have learned how to deal with series in complete normed vector spaces, continuous maps between arbitrary complete normed vector spaces, continuous linear maps between normed vector spaces and the shrinking lemma.

• Have you proved that $\log(\exp(A)) = \exp(\log(A)) = A$? Dec 3, 2020 at 20:12
• Ah, I just realized that the proof follows directly from that fact. I hadn't proved it since the proof requires Taylor series error estimates, which we haven't learned about formally. Dec 3, 2020 at 20:29

Using the fact that $$\log(\exp(A))=\exp(\log(A))=A$$, we have $$\exp(\log(A)+\log(B))=\exp(\log(A))\exp(\log(B))$$ which reduces to $$\exp(\log(A)+\log(B))=AB.$$ If we take the Log of both sides we have $$\log(\exp(\log(A)+\log(B)))=\log(AB)$$ and hence $$\log(A)+\log(B)=\log(AB)$$