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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?

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    $\begingroup$ Good first steps are: 1) identify the eigenvalues of $A$ 2) use the fact that $A$ is diagonalizable iff its minimum polynomial is a product of distinct linear factors. $\endgroup$ – Matthew Towers Jul 21 '16 at 13:36
  • $\begingroup$ Thanks. Point (1) is easy. But is there an easy way to see the minimum polynomial? $\endgroup$ – W. Rether Jul 21 '16 at 13:43
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    $\begingroup$ The eigenvalues are the roots of the min poly. So if you know what the eigenvalues are, you know what the min poly must be in the diagonalizable case. Now test which $A$ actually satisfy that "min poly". $\endgroup$ – Matthew Towers Jul 21 '16 at 13:44
  • $\begingroup$ Thank you. Problem solved. $\endgroup$ – W. Rether Jul 21 '16 at 13:50
  • $\begingroup$ please write a solution here, I'd be interested to see it $\endgroup$ – Matthew Towers Jul 21 '16 at 13:52
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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:

  1. 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, $c_d = c_{d-1} = \dots = c_0 = 0$.
  2. 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:

  1. 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 $A = 0$.
  2. 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 $c_{d-1} = \dots = c_0 = 0$.
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    $\begingroup$ I've written this solution mostly to show how the explicit calculation of the product $A(A - c_d I)$ can be avoided. $\endgroup$ – levap Jul 21 '16 at 14:32

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