Show the existence of the invertible matrix. It's a problem from the book "Differential Equations,Dynamical Systems,and Linear Algebra" by Hirsch and Smale:

Let $A$ be a $2\times 2$ matrix whose eigenvalues are the complex numbers $\alpha\pm \beta i$,$\beta\ne 0$.Let $B=\left[ \begin{matrix}
 \alpha&  -\beta\\
 \beta&  \alpha\\
\end{matrix} \right]$.
Show there exists an invertible matrix $Q$ with $QAQ^{-1}=B$,as follows:
(a).Show that the determinant of the following $4\times 4$ matrix is $0$:
$$
\left[ \begin{matrix}
 A-\alpha I&  -\beta I\\
 \beta I&  A-\alpha I\\
\end{matrix} \right] ,
$$
(b).Show that there exists a $2\times 2$ matrix $Q$ such that $AQ=QB$.(Hint: Write out the above equation in the four entries of $Q=[q_{ij}]$ Show that the resulting system of four-linear homogeneous equations in the four unknowns $q_{ij}$ has the coefficient matrix of part (a).)
(c) Show that $Q$ can be chosen invertible.

Part (a) and (b) are easy to prove,but I don't know how to tackle with (c). It seems that the rank of the matrix in (a) is 3, hence the solution space of the corresponding four-linear homogeneous equations is just consisted by 1 base. How can we say that the base vector $(q_1,q_2,q_3,q_4)$ satisfies that $q_1\times q_4\ne q_2\times q_3.$
 A: Write $Q = [q_1 \, | \, q_2]$ where $q_1,q_2$ are the columns of $Q$. Then $Q$ satisfies $AQ = QB$ if and only if we have
$$ \begin{bmatrix} A - \alpha I & - \beta I \\ \beta I & A - \alpha I \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} = \begin{bmatrix} 0_{2 \times 1} \\ 0_{2 \times 1} \end{bmatrix}. $$
This gives you the two equations:
$$ (A - \alpha I) q_1 = \beta q_2, \,\,\, (A - \alpha I)q_2 = -\beta q_1. $$
Let $Q$ be a non-zero solution of $AQ = QB$ and let's show that $Q$ must be invertible. Assume that $Q$ is not invertible. Then we have two cases:

*

*If $q_1 \neq 0$ then $q_2 = \lambda q_1$ for $\lambda \in \mathbb{R}$. But then $(A - \alpha I)(q_1) = \lambda \beta q_1$ which means that $Aq_1 = (\alpha + \lambda \beta)q_1$ so $q_1$ is an eigenvector associated to the real eigenvalue $\alpha + \lambda \beta$. But $A$ already has two distinct non-real eigenvalues, a contradiction.

*If $q_2 \neq 0$ then $q_1 = \lambda q_2$ for $\lambda \in \mathbb{R}$. But then $(A - \alpha I)(q_2) = -\lambda \beta q_2$ which means that $Aq_2 = (\alpha - \lambda \beta)q_2$ and again $q_2$ is an eigenvector associated to the real eigenvalue $\alpha - \lambda \beta$, a contradiction.

