Why is it not true that $e^{X} \oplus 1 = e^{X \oplus 1}$?

Question

Consider a direct sum vector space $V = V_1 \oplus V_2$, and let $X$ be a matrix acting on $V_1$. I recently derived the equation in the title of this question, where $1$ denotes identity matrix on $V_2$, in the following way. Since $$1 = 1^n$$ for all $n \geq 0$, and, $$AB \oplus CD = (A\oplus C)(B\oplus D)$$ for all $A,B,C,D$, we have: \begin{align} e^{X}\oplus 1 &= \left(\sum_{n=0}^{\infty}\frac{X^n}{n!}\right)\oplus 1\\ &= \sum_{n=0}^{\infty} \left(\frac{X^n}{n!} \oplus 1^n\right)\\ &= \sum_{n=0}^{\infty} \frac{(X \oplus 1)^n}{n!}\\ &= e^{X \oplus 1}. \end{align} However, I tried to verify this equation numerically using Python, and I found that it is actually false. Could someone help me identify the error in my derivation?

Answer 1

The issue was correctly pointed out by @anomaly in the second equality, but maybe an example will help.

Let $X = \begin{bmatrix} 2\end{bmatrix}$ and $1=I_1=\begin{bmatrix} 1\end{bmatrix}$. Then $e^X = \begin{bmatrix}e^2\end{bmatrix}$ and $e^X\oplus 1 =\begin{bmatrix}e^2 & 0 \\ 0 & 1 \end{bmatrix}$, while $X\oplus 1=\begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}$ which has the exponential $e^{X\oplus 1}=\begin{bmatrix}e^2 & 0 \\ 0 & e^1 \end{bmatrix}$.

Essentially you use an identity of the form

$(A+A')\oplus B = (A\oplus B)+(A'\oplus B)$ but this isn't true in general as the right hand side would have a copy of $2B$ in the bottom right. As pointed out in the comments, it is true if $B=0$!