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I have to determine the characteristic polynomial of the matrix $$A = \begin{pmatrix} 0 & 0 &\cdots &0& -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & \cdots & 0&-a_2 \\\vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 &\cdots & 1 & -a_{n-1} \end{pmatrix} . $$

I tried to achieve a triangular matrix $A^*$ before applying $\vert\lambda I - A)\vert$ because then the determinant is very easy to calculate; this yielded the characteristic polynomial $(\lambda - 1)^{n-1} * (\lambda + a_{n-2})$. However I learnt afterwards that this approach is erroneous and the characteristic polynomial of $A^*$ would not reflect that of the original matrix $A$. What is the right approach?

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  • $\begingroup$ One correct approach is just to find the determinant of $\lambda I - A$. Do it explicitly for $n=1, 2, 3, 4$ and see if you notice a pattern. Then prove that that pattern holds in general. $\endgroup$ – Alex G. May 21 '16 at 15:52
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    $\begingroup$ This is called a Frobenius companion matrix, and the characteristic polynomial is a monic polynomial with lower order coefficients given by the $a_i$'s. $\endgroup$ – hardmath May 21 '16 at 20:54
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Hint: For $n=1$, $\det(A-tI)=-t-a_0$. For $n=2$, $\det(A-tI)=t^2+a_1t+a_0$. For $n=3$, $\det(A-tI)=-t^3-a_2t^2-a_1t-a_0$. So one may claim that in general case, $$\det(A-tI)=(-1)^n\left(t^n+\displaystyle\sum_{i=0}^{n-1}a_it^i\right),$$ by using the induction on $n$.

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