Q:

Prove that $\theta$ is automorphism

Let $V$ be a vector space over a field $F$; let $\alpha \in End(V )$ and let $\beta \in Aut(V )$. Define the function $\theta: V^2\to V^2$ by setting $\theta:\begin{bmatrix} v \\ v'\end{bmatrix}\to\begin{bmatrix} \beta(v) \\ \alpha(v)+v'\end{bmatrix}$. Is $\theta$ necessarily an automorphism of $V^2$?
I know $\theta$ is a linear transformation. I tried to make an argument to prove $theta$ is bijective, but I'm not sure about it. 
Let $\begin{bmatrix} v \\ v'\end{bmatrix}\in Ker(\theta)$, then we have $\begin{bmatrix} \beta(v) \\ \alpha(v)+v'\end{bmatrix}=\begin{bmatrix} 0\\0\end{bmatrix}$. Since $\beta\in Aut(V)$, it is monic, so, $v=0$.
This implies $\alpha(v)=0$, and then $v'=0$. 
To show that $\theta$ is epic, let $\begin{bmatrix}y\\w\end{bmatrix}\in V^2$. As, $\beta $ is an automorphism there exists $v\in V$ satisfying $\beta(v)=y$. For such $v$, it's enough to take $v'=w-\alpha(v)$. Hence, $\theta$ is bijective, since we could find $\begin{bmatrix}v\\v'\end{bmatrix}$ satisfying $\theta\Big(\begin{bmatrix}v\\v'\end{bmatrix}\Big)=\begin{bmatrix}y\\w\end{bmatrix}$ .

A:

Yes, the proof is correct. You might also consider the “matrix” of $\theta$:
$$
\begin{bmatrix}
\beta & 0 \\
\alpha & \iota
\end{bmatrix}
$$
where $\iota$ is the identity. This is a matrix over the endomorphism ring of $V$, which has an inverse
$$
\begin{bmatrix}
\beta & 0 \\
\alpha & \iota
\end{bmatrix}
\begin{bmatrix}
\beta^{-1} & 0 \\
-\alpha\beta^{-1} & \iota
\end{bmatrix}=
\begin{bmatrix} \iota & 0 \\ 0 & \iota\end{bmatrix}
$$
and the same switching sides, which essentially tells you that
$$
\theta^{-1}\begin{bmatrix}v\\v'\end{bmatrix}=
\begin{bmatrix} \beta^{-1}(v)\\v'-\alpha\beta^{-1}(v)\end{bmatrix}
$$
(that you basically discovered too).

