Let $A$ be an $N \times N$ real (not necessarily symmetric) matrix with distinct eigenvalues. I want to diagonalise this over the reals. Of course, you can't always do this but I have seen it claimed that you can block-diagonalise in the following way. Let $\lambda_1, \dots , \lambda_R$ be the real eigenvalues of $A$ and $\mu_1, \overline{\mu}_1, \dots, \mu_Q, \overline{\mu}_Q$ be the strictly complex eigenvalues of $A$, where $R + 2Q = N$. Let $\mu_k = a_k + i b_k$ , $a_k, b_k \in \mathbb{R}$.
Then there is an invertible real transformation (change of basis) that puts $A$ in the form $$\mathrm{diag}(\lambda_1, \dots,\lambda_R) \oplus \left( \begin{matrix} a_1 & b_1 \\ -b_1 & a_1 \end{matrix} \right) \oplus \dots \oplus \left( \begin{matrix} a_Q & b_Q \\ -b_Q & a_Q \end{matrix} \right) $$ How can one show this? I have also seen a claim (which I assume is related) that any $2 \times 2$ real matrix can be put in the form $\left( \begin{matrix} a & b \\ -b & c \end{matrix} \right)$ by an orthogonal transformation. I would also like to know how to prove that.
For the second claim, the proof goes as follows. Let $A$ be a $2 \times 2$ real matrix. Then one may write $A = \frac{A+A^T}{2} + \frac{A-A^T}{2}$. There is an orthogonal transformation $V$ that diagonalises the symmetric part $$V\frac{A+A^T}{2}V^T = \left( \begin{matrix} a & 0 \\ 0 & c \end{matrix} \right)$$. Anti-symmetry is preserved under this transformation, so $$V\frac{A-A^T}{2}V^T = \left( \begin{matrix} 0 & b \\ -b & 0 \end{matrix} \right)$$ which completes the proof.