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I know that is possible to apply the spectral decomposition (diagonalization) to a matrix when the sum of the dimensions of its eigenspaces is equal to the size of the matrix.

The spectral decomposition is:

$$ F=P\Lambda P^{-1} $$

where $\Lambda$ is the diagonal matrix of eigenvalues and $P$ is the matrix of eigenvectors.

I have the following matrix:

$$ F=\begin{pmatrix}\phi_{1} & \phi_{2} & \phi_{3} & \cdots & \phi_{p-1} & \phi_{p}\\ 1 & 0 & 0 & \cdots & 0 & 0\\ 0 & 1 & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & \cdots & 1 & 0 \end{pmatrix} $$

where $\phi_{p}\neq0$ and all $\phi_{i}$ are real valued. How can I be sure that its possible to apply the spectral decomposition to $F$?

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  • $\begingroup$ Are the $\phi_j$ complex? $\endgroup$ – Aryaman Jal Nov 25 '17 at 14:08
  • $\begingroup$ no, they are real valued. $\endgroup$ – Gustavo Amarante Nov 25 '17 at 14:08
  • $\begingroup$ That matrix is not always diagonalizable, so you can't apply the spectral decomposition $\endgroup$ – Exodd Nov 25 '17 at 14:13
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The matrix you wrote is the Companion Matrix of the polynomial $$ P(x) = -( \phi_p + x\phi_{p-1} + \dots + x^{p-1}\phi_1 + x^p ) $$ It is known that the matrix is diagonalizable if and only if $P(x)$ has all $p$ different roots. That can be also easily proved through the test of the derivative:

$P(x)$ has multiple roots $\iff$ $gcd(P(x),P'(x))\ne 1$

where $P'(x)$ is the derivative of $P(x)$.

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A complex matrix is diagonalizable iff the minimal polynomial has distinct roots. The minimal polynomial of your matrix is $x^p+\phi_1 x^{p-1}+\cdots+\phi_{n-1}x+\phi_{n}$. So, you know that your matrix is not always diagonalizable.

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