I would like to know whether the diagonalizable matrices is dense in the set of matrices which is block diagonal with companion matrix in each block and the columns of each block extended to the whole matrix. If this is not clear, I will demonstrate by an example.
To make it concrete, let us consider an example \begin{align*} A = \begin{pmatrix} 0 & -a_1 & 0 & 0 & -b_1 \\ 1 & -a_2 & 0 & 0 & -b_2 \\ 0 & -a_3 & 0 & 0 & -b_3 \\ 0 & -a_4 & 1 & 0 & -b_4 \\ 0 & -a_5 & 0 &1 & -b_5 \end{pmatrix}. \end{align*} Let $\mathcal E \subset \mathcal M(5 \times 5; \mathbb R)$ be the collection matrices in above form and $\mathcal D \subset \mathcal M(5 \times 5; \mathbb R)$ be the set of matrices having distinct eigenvalues in $\mathbb C$. We know $\mathcal D$ is dense in $\mathcal M(5 \times 5; \mathbb R)$. I wonder whether $\mathcal D \cap \mathcal E$ is dense in $\mathcal E$.
Let $F: \mathcal E \to \mathbb R^5$ be the map of extracting the coefficients of characteristic polynomial of $\mathcal E$. That is, for $A \in \mathcal E$, $F$ is defined by $A \mapsto (\alpha_4, \alpha_3, \dots, \alpha_0)$ where $p_A(t) = t^5 + \alpha_4 t^4 + \dots + \alpha_0$. We know the polynomials with distinct roots in dense in monic polynomials of order $5$. So if the map $F$ is open, we may conclude by noting $\mathcal D \cap \mathcal E$ is the inverse image of a dense set. But I don't know whether $F$ is open and can only see $F$ is closed. It could be the statement is false.
The characteristic polynomial of your $A$ is $$ {\lambda}^{5}- \left( -b_{{5}}-a_{{2}} \right) {\lambda}^{4}- \left( - b_{{5}}a_{{2}}+a_{{5}}b_{{2}}-a_{{1}}-b_{{4}} \right) {\lambda}^{3}- \left( -b_{{5}}a_{{1}}-a_{{2}}b_{{4}}+a_{{4}}b_{{2}}+a_{{5}}b_{{1}}-b _{{3}} \right) {\lambda}^{2}- \left( -a_{{1}}b_{{4}}-a_{{2}}b_{{3}}+a_ {{3}}b_{{2}}+a_{{4}}b_{{1}} \right) \lambda+a_{{1}}b_{{3}}-a_{{3}}b_{{ 1}} $$ For e.g. $a_1 = 1,\; a_2 = 2,\; \ldots, b_5 = 10$ this is $$ {\lambda}^{5}+12\,{\lambda}^{4}-5\,{\lambda}^{3}-22\,{\lambda}^{2}-20 \,\lambda-10 $$ which has discriminant $-69862372720 \ne 0$. Since the discriminant is a polynomial in $a_1, \ldots, b_5$, this is nonzero for almost every $a_1, \ldots, b_5$. When the discriminant is nonzero, the eigenvalues are distinct and the matrix is diagonalizable.