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Let $$A = \begin{bmatrix} a_1 & a_2 & a_3 & \cdots & a_{n-1} & a_n\\ 1& 0 & 0 & \cdots & 0 & 0\\ 0 & 1 & 0 & \cdots & 0 &0 \\ \vdots & \vdots & \ddots & \ddots & \vdots &\vdots \\ 0& 0& 0 &\ddots &0 &0 \\ 0&0 &0 & \cdots & 1 & 0 \end{bmatrix}.$$ What is the characteristic polynomial of $A$?

for $n = 2$:

$$A = \begin{bmatrix} a_1 & a_2\\ 1& 0 \end{bmatrix} \Rightarrow \det(A - \lambda I) = -\lambda(a_1 - \lambda) - a_2,$$ which is equivalent to $$\det(A - \lambda I) = (-1)^{2-1}\left[\lambda^{2-1}(a_1 - \lambda) + \lambda^{2-2}a_2\right].$$

for $n = 3$: $$A = \begin{bmatrix} a_1 & a_2 & a_3\\ 1&0 &0 \\ 0&1 &0 \end{bmatrix} \Rightarrow \det(A - \lambda I) = (-1)^2[\lambda^2(a_1 - \lambda) + \lambda a_2 + a_3],$$ which is equivalent to $$\det(A - \lambda I) = (-1)^{3-1}[\lambda^{3-1}(a_1 - \lambda) + \lambda^{3-2} a_2 + \lambda^{3-3}a_3].$$

For $n = 4$ $$A = \begin{bmatrix} a_1 &a_2 &a_3 &a_4 \\ 1&0 &0 &0 \\ 0&1 &0 &0 \\ 0 &0 &1 &0 \end{bmatrix} \Rightarrow \det(A - \lambda I ) = (-1)^3[\lambda^3(a_1-\lambda) + \lambda^2 a_2 + \lambda a_3 + a_4],$$ which is equivalent to $$\det(A - \lambda I ) = (-1)^{4-1}[\lambda^{4-1}(a_1-\lambda) + \lambda^{4-2} a_2 + \lambda^{4-3} a_3 + \lambda^{4-4}a_4].$$

It seems that in general for $n = k$, we have: $$\det(A - \lambda I) = (-1)^{k-1}[\lambda^{k-1}(a_1 - \lambda) + \lambda^{k-2}a_2 + \lambda^{k-3}a_3 + \cdots + \lambda a_{k-1} + a_k].$$

Is this argument the right proof of the problem?

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  • $\begingroup$ It seem correct. The characteristic polynomial should be - up to a sign - $\lambda^k - \sum_{i=1}^{k}a_{i}\lambda^{k-i}$. $\endgroup$ – TZakrevskiy Sep 20 '16 at 18:33
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    $\begingroup$ Keyword companion matrix $\endgroup$ – quid Sep 20 '16 at 18:33
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Instead of explicitly computing the characteristic polynomial, one might argue as follows: $ A $ and $ A^T $ have the same characteristic polynomial, so let $ B = A^T $, let $ e_k $ denote the elementary basis vectors and note the following:

$$ B e_1 = \sum_{k=1}^n a_k e_k $$

Now, since $ B^m e_n = e_{n-m} $ for $ m < n $, the vectors $ e_n, B e_n, \ldots, B^{n-1} e_n $ are linearly independent, and hence the minimal polynomial of $ B $ has degree $ n $, and is equal to its characteristic polynomial. Now, rearranging the equality yields $ p(B) e_n = 0 $, where

$$ p(X) = X^n - a_1 X^{n-1} - \ldots - a_n $$

and it is then easily seen that $ p(B) $ kills all of the elementary basis vectors, and hence kills $ B $ as a linear operator. Since we know that the minimal polynomial has degree $ n $, it follows that the minimal polynomial, and therefore the characteristic polynomial, is $ p(X) $.

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