I'm reading through the book Gauge Fields, Knots, and Gravity by John Baez, and trying to make sure I have a firm grasp on gauge transformations. To that end, I've been looking at his concrete example of a gauge transformation of the trivial $G$-bundle $E=M\times\mathbb{C}$ where $G = U(1)$. The example goes as follows:
[NOTE: This book is a book on mathematics as it applies to physics and as such there is a lot of "sloppiness", for example many times when dealing with representations of a group, I will write $g$ when really I mean $\rho(g)$, so please bear with me.]
The connection $D$ on E can be described by the vector potential $A$, an End($E$)-valued 1-form. But since we have a trivial bundle $E=M\times\mathbb{C}$, End($E$) = End($\mathbb{C}$) $= \mathbb{C}$. But if we think of the connecton $D$ as a $U(1)$ connection, we want the components, $A_\mu$ of the vector potential to live in $\mathfrak{u}(1)$, i.e., we want $A_\mu$ to be of the form $A_\mu(p,v)\longrightarrow(p,d\rho(x)v)$ where $\rho$ is the representation of $U(1)$ on $\mathbb{C}$. Since $\mathfrak{u}(1) = \{ix\ |\ x\in\mathbb{R}\}$, and $\rho$ is really just the map $\rho:U(1)\longrightarrow U(1)\subset\mathbb{C}$, this amounts to the $A_\mu$'s just being imaginary functions.
Now, we apply a gauge transformation $g$: $$A_\mu' = gA_\mu g^{-1} + g\partial_\mu g^{-1}$$ But $U(1)$ is abelian, so we get $$A_\mu' = A_\mu + g\partial_\mu g^{-1}$$ and since $g$ is a gauge transformation, it lives in $U(1)$ and so has the form $g(p)(p,v)=(p,\rho(g(p)) v)$ we can write $$g = e^{-f(p)}$$ so that $$A_\mu' = A_\mu + \partial_\mu f$$ or $$A' = A + df$$which is the usual result from electromagnetism.
Now, to make sure I understand this stuff correctly I wanted to go through the same argument on my own using a different representation of $U(1)$, namely $\rho:U(1) \longrightarrow SO(2) \subset GL(\mathbb{R}^2)$. In this case, we take the $G$-bundle to be the trivial bundle $E = M\times\mathbb{R}^2$ and we want $A_\mu$ to live in $\mathfrak{u}(1)$ which means it has the form $A_\mu(p,v)\longrightarrow(p,d\rho(x)v)$ where $x\in\mathfrak{u}(1)$ and $d\rho(x)$ is now an element of $\mathfrak{so}(2)$ and has the form $d\rho(i\theta)=\theta\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. So, really the $A_\mu$'s are real $2\times 2$ matrix-valued functions on $M$. We apply a gauge transformation, $g$ as before and obtain $$A_\mu'=A_\mu + g\partial_\mu g.$$ $g$ lives in $U(1)$ and so has the form $g(p)(p,v)\longrightarrow(p,\rho(g(p))v)$ so we can write $$g(p) = \left(\begin{matrix}\cos(f(p))&\sin(f(p))\\ -\sin(f(p))&\cos(f(p))\end{matrix}\right)$$ which gives $$A_\mu'=A_\mu + \left(\begin{matrix}0&-1\\1&0\end{matrix}\right)\partial_\mu f$$ Now, I realize this looks messy since $A_\mu$ are really matrices themselves so I have $$\left(\begin{matrix}0&-(A_\mu')_{12}\\(A_\mu')_{21}&0\end{matrix}\right) = \left(\begin{matrix}0&-(A_\mu)_{12}\\(A_\mu)_{21}&0\end{matrix}\right) + \left(\begin{matrix}0&-\partial_\mu f\\\partial_\mu f&0\end{matrix}\right)$$
And here is where I am not sure if I am right: since $A_\mu$ is of the form $\theta\left(\begin{matrix}0&-1\\1&0\end{matrix}\right)$ then it must be anti-Hermitian so that $(A_\mu)_{12} = (A_\mu)_{21}$? If this is right, then I can think of the components of the vector potential as a real-valued functions on $M$ that satisfy $A_\mu' = A_\mu + \partial_\mu f$ as expected. Is this a valid argument?