# Any proof of $(E+A+\frac{1}{2!}A^2+\cdots)(E+B+\frac{1}{2!}B^2+\cdots)=E+(A+B)+\frac{1}{2!}(A+B)^2+\cdots$ which doesn't use norm of matrices?

I am reading "Linear Algebra" by Ichiro Satake.

Let $$A_0,A_1,\dots$$ be a sequence of complex $$n\times n$$ matrices.
We define $$A_0+A_1+\cdots$$ converges if $$a_{ij}^{(0)}+a_{ij}^{(1)}+\cdots$$ converges for any $$i,j\in\{1,\dots,n\}$$, where $$a_{ij}^{(\nu)}$$ is the $$(i,j)$$ element of $$A_{\nu}$$.

Let $$A$$ be a complex $$n\times n$$ matrix.
Then, $$E+A+\frac{1}{2!}A^2+\cdots$$ converges.

Let $$A$$ be a complex $$n\times n$$ matrix.
Let $$B$$ be a complex $$n\times n$$ matrix.
Suppose that $$AB=BA$$.
Then, $$(E+A+\frac{1}{2!}A^2+\cdots)(E+B+\frac{1}{2!}B^2+\cdots)=E+(A+B)+\frac{1}{2!}(A+B)^2+\cdots$$ holds.

Satake says the proof of $$(E+A+\frac{1}{2!}A^2+\cdots)(E+B+\frac{1}{2!}B^2+\cdots)=E+(A+B)+\frac{1}{2!}(A+B)^2+\cdots$$ is the same as the proof of $$e^ae^b=e^{a+b}$$.

But Satake doesn't define norm of a matrix in this book.

Is there any elementary proof which doesn't use norm of a matrix to prove the above equation?

• Ea... Since $AB=BA$ and as the hint shows: let $f(x)=1+x+1/2x^2+\dots$ be the Taylor expansion of $e^x$ then $e^a\times e^b=e^{a+b}$. This identity can be proven by using the binomial coefficients and expanding $(a+b)^k$. You can do the analogous thing to the matrix that commute by multiplication. Commented Sep 5 at 2:13
• @JetfiRex thank you very much. Commented Sep 5 at 8:46