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I am trying to find the characteristic polynomial of: $$ A= \begin{pmatrix} \alpha_1 & \alpha_2 & \cdots & \alpha_{55} & \\ \alpha_1 & \alpha_2 & \cdots & \alpha_{55} & \\ \vdots & \vdots & \ddots & \vdots & \\ \alpha_1 & \alpha_2 & \cdots & \alpha_{55} & \end{pmatrix} $$
$\forall i: 1\leq i\leq 55: a_i$ are scalars.

I didn't actually got the trick here, any advice?

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  • $\begingroup$ Thank you for your hint and comment, I appreciate it! $\endgroup$ – user86419 Jul 15 '13 at 21:49
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$$ \det\begin{pmatrix} \alpha_1-\lambda & \alpha_2 & \cdots & \alpha_{55} & \\ \alpha_1 & \alpha_2-\lambda & \cdots & \alpha_{55} & \\ . & . & . & . & \\ .& . & . & . & \\ \alpha_1 & \alpha_2 & \cdots & \alpha_{55}-\lambda & \end{pmatrix} $$

Substract the first row from every other row

$$ =\det\begin{pmatrix} \alpha_1-\lambda & \alpha_2 & \cdots & \alpha_{55} & \\ \lambda & -\lambda & \cdots & 0 & \\ . & . & . & . & \\ .& . & . & . & \\ \lambda & 0 & \cdots & -\lambda & \end{pmatrix} $$

Compute the transpose

$$ =\det\begin{pmatrix} \alpha_1-\lambda & \lambda & \cdots & \lambda & \\ \alpha_2 & -\lambda & \cdots & 0 & \\ . & . & . & . & \\ .& . & . & . & \\ \alpha_{55} & 0 & \cdots & -\lambda & \end{pmatrix} $$

Add every row to the first one

$$ =\det\begin{pmatrix} \alpha_1+\cdots +\alpha_{55}-\lambda & 0 & \cdots & 0 & \\ \alpha_2 & -\lambda & \cdots & 0 & \\ . & . & . & . & \\ .& . & . & . & \\ \alpha_{55} & 0 & \cdots & -\lambda & \end{pmatrix} $$

Therefore you get $\lambda^{54}(\alpha -\lambda)$ where $\alpha=\alpha_1+\cdots +\alpha_{55}$.

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  • $\begingroup$ This one I like. $\endgroup$ – Git Gud Jul 15 '13 at 22:15
  • $\begingroup$ I also like this answer, it covers the case $\alpha=0$. Thank you all for your time and effort! I really appreciate it you helped me a lot! thanks! $\endgroup$ – user86419 Jul 15 '13 at 22:23
  • $\begingroup$ You're welcome. :) $\endgroup$ – ՃՃՃ Jul 15 '13 at 22:25
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The case $\alpha_i=0\ \forall i$ is trivial so let's treat the general case

The rank of the matrix $A$ is $1$ so by the rank nullity theorem $\dim \ker A=54$ and then $0$ is an eigenvalue of $A$ with multiplicity $54$ and the last eigenvalue is $\displaystyle \mathrm{tr}(A)=\sum_{k=1}^{55}\alpha_i=\lambda$ and then $$\chi_A(x)=x^{54}(x-\lambda)$$

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  • $\begingroup$ I'm confused about something. How do you conclude that $\lambda \neq 0$? Or if you allow it to be $0$, how do you make sure you're not already counting it in the rank nullity theorem? $\endgroup$ – Git Gud Jul 15 '13 at 21:30
  • $\begingroup$ $\lambda$ my be $0$ for example take $\alpha_1=1$ and $\alpha_2=-1$ and the other $\alpha_i=0$. $\endgroup$ – user63181 Jul 15 '13 at 21:35
  • $\begingroup$ Yes and my second question? $\endgroup$ – Git Gud Jul 15 '13 at 21:35
  • $\begingroup$ Thank you both for your answers! @GitGud But what happens if all α 1 +...+α 55 =0 ? $\endgroup$ – user86419 Jul 15 '13 at 21:38
  • $\begingroup$ I don't well understand this question but the only matrix with $\dim\ker A=55$ is the zero matrix. $\endgroup$ – user63181 Jul 15 '13 at 21:39

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