Diagonalizability of a given matrix I must find out under which conditions the matrix
$$A=
    \left[\begin{array}{ccc|cc}& & & c_0 &\\
    &  & &c_1&\ddots\\
& & &c_2 &\ddots& c_0\\
& 0 & & \vdots & \ddots &c_1\\
& & & c_d & \ddots &c_2\\
& & & & \ddots & \vdots\\ 
& & & & & c_d\\
& & &  &&   \end{array}\right]
$$
is diagonalizable (where the $c_d$ lie on the diagonal and the first $d$ columns are $0$, and so everywhere else).
Perhaps the Jordan form is easy to calculate?

Note that your matrix $A$ is a block upper triangular matrix of the form
$$ A = \begin{pmatrix} 0_{d \times d} & B \\ 0 & D_{(d + 1) \times (d + 1)} \end{pmatrix} $$
where $D$ is also a block upper triangular matrix with $c_d$ on the diagonal. Hence, the characteristic polynomial of $A$ is
$$ p_A(x) = x^d(x - c_d)^{d+1}. $$
We have two options:


*

*If $c_d = 0$ then the characteristic polynomial of $A$ is $x^{2d + 1}$ and so $A$ is nilpotent which implies that if $A$ is also diagonalizable we must have $A = 0$. Hence, $c_d = c_{d-1} = \dots = c_0 = 0$.

*If $c_d \neq 0$ then $A$ will be diagonalizable if and only if the minimal polynomial of $A$ is $x(x - c_d)$. Plugging $A$ into $x(x - c_d)$, we see that we must have
$$ A(A - c_d I) = \begin{pmatrix} 0 & B \\ 0 & D \end{pmatrix} 
\begin{pmatrix} -c_d I & B \\ 0 & D - c_d I \end{pmatrix} = \begin{pmatrix} 0 & B(D - c_d I) \\ 0 & D(D - c_d I) \end{pmatrix} = 0. $$
In particular, $D(D - c_dI) = 0$ and so $D$ must be diagonalizable with a single eigenvalue $c_d$ which implies that in fact $D = c_dI$ and so $c_{d-1} = \dots = c_0 = 0$.
In any case, we see that we must have $c_{d-1} = \dots = c_0$ for the matrix to be diagonalizable.

An alternative, more elementary, solution that doesn't involve the minimal polynomial continuous from the calculation of the characteristic polynomial as follows:


*

*If $c_d = 0$ then the characteristic polynomial of $A$ is $x^{2d+1}$ and so if $A$ is diagonalizable, we must have $\dim \ker(A) = 2d + 1$ so $A = 0$.

*If $c_d \neq 0$ then we must have $\dim \ker A = d$ and $\dim \ker (A - c_d I) = d + 1$. Since $c_d \neq 0$, the $d + 1$ non-zero columns of $A$ are linearly independent and so $\dim \ker A = d$. However,
$$ A - c_dI = \begin{pmatrix} -c_dI & B \\ 0 & D - c_dI \end{pmatrix} $$
and so the first $d$ columns of $A - c_dI$ are linearly independent. In order that $\dim \operatorname{Im}(A - c_dI) = d$, the rest of the columns must belong to the span of the first $d$ columns. By looking at the last column, we see the this implies that $c_{d-1} = \dots = c_0 = 0$.

