Background
I am solving an assignment on the fundamentals of Quantum Computing which mostly has questions on Linear Algebra. All matrices and vectors are over complex domains.
Question
Define the exponent of a matrix as $$e^{\lambda A} = \sum_{i = 0}^{\infty}\frac{\lambda^i}{i!}A^i$$ where $A^0 = \mathbb{I}$ or the identity operator.
- Given a matrix $B$, the matrix logarithm of $B$ is defined as the matrix $A$ such that $e^A = B$. Prove that $\log$ is a non-unique function (i.e. for every $B$, there are several $A$ that satisfy the matrix formula given).
- Prove that for two matrices, if $AB = BA$, then $\log(AB) = \log(A) + \log(B)$.
My approach
- For this part, I am firstly not sure about the existence of an $A$, it doesn't seem obvious to me why such an $A$ should always exist and if so, how to compute it. Nonetheless, assuming that we have a solution to $\log(B) = A$, then I try to prove that I can derive a family of solutions from it (this is my intuition) $$\therefore B = \mathbb{I} + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \ldots $$ Assume we have an $A' = A + 2\pi i \mathbb{I}$ such that $$ e^{A'} = \mathbb{I} + (A+2\pi i)\mathbb{I} + \frac{({A + 2\pi i\mathbb{I}})^2}{2!} + \frac{({A + 2\pi i\mathbb{I}})^3}{3!} + \ldots $$ Since $A$ and $\mathbb{I}$ always commute, we can write the sum as (the proof is just algebraic manipulation and application of the binomial theorem, so I've skipped it) $$e^{A'} = \left(\mathbb{I} + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \ldots\right)\left(\mathbb{I} + \frac{2\pi i}{1!}\mathbb{I} + \frac{(2\pi i)^2}{2!}\mathbb{I} + \frac{(2 \pi i)^3}{3!} \mathbb{I} + \ldots \right)$$ $$\therefore e^{A'} = e^A \mathbb{I} \left(1 + \frac{2\pi i}{1!}+ \frac{(2\pi i)^2}{2!}\ + \frac{(2 \pi i)^3}{3!} + \ldots \right)$$ $$\therefore e^{A'} = e^A \mathbb{I} e^{2 \pi i} = B$$ Similarly $\forall n \in \mathbb{Z}, \exp{(A + 2n \pi i \mathbb{I})} = B$ This finishes the proof
- Assume $e^X = AB, e^Y = A, e^Z = B$ To prove that $X = Y + Z$ We write out the expansions of $A$ and $B$ and then use the fact that $AB = BA$ $$AB = \mathbb{I} + (Y + Z) + \frac{Y^2 + 2YZ + Z^2}{2!} + \frac{Y^3 + 3Y^2Z + 3YZ^2 + Z^3}{3!} + \ldots$$ $$BA = \mathbb{I} + (Y + Z) + \frac{Y^2 + 2ZY + Z^2}{2!} + \frac{Y^3 + 3ZY^2 + 3Z^2Y + Z^3}{3!} + \ldots$$ Put $AB - BA = \mathbb{O}$ $$\therefore (YZ - ZY) + \frac{1}{2!} (Y^2Z - ZY^2) + \frac{1}{2!}(YZ^2 - Z^2Y) + \ldots = 0$$
This is where I am stuck Obviously if $Y$ and $Z$ commute, then the above expression equates to 0 and thus from the expansions of $AB$ and $BA$, if $Y$ and $Z$ would commute, $AB = BA = e^{Y+Z} = e^X$ and thus our proof is complete.
The question is: how do I prove that this is the only case where the summation is 0? If $Y$ and $Z$ don't commute, then is it possible to proceed with the proof? Are there any alternate approaches?
