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The question is to find the minimum polynomial of the following n\times n matrix: $A=\begin{pmatrix}1 & 1 & \cdots & 1\\ 1 & \ddots & \ddots & 1\\ \vdots & \ddots & \ddots & \vdots\\ 1 & 1 & \ddots & 1 \end{pmatrix}$ , $n\geq2$.

This is what I've tried:

I calculate the characteristic polynomial as follows:

$$p_{A}(x)=\left|\begin{array}{cccc} 1-x & 1 & \cdots & 1\\ 1 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & 1\\ 1 & \cdots & 1 & 1-x \end{array}\right| =\left|\begin{array}{cccc} n-x & n-x & \cdots & n-x\\ 1 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & 1\\ 1 & \cdots & 1 & 1-x \end{array}\right|=(n-x)\left|\begin{array}{cccc} 1 & 1 & \cdots & 1\\ 1 & 1-x & \cdots & \vdots\\ \vdots & \ddots & \ddots & 1\\ 1 & \cdots & 1 & 1-x \end{array}\right|=(n-x)\left|\begin{array}{cccc} 1 & 0 & \cdots & 0\\ 1 & -x & \cdots & \vdots\\ \vdots & \ddots & \ddots & 0\\ 1 & 0\cdots & 0 & -x \end{array}\right|=(n-x)(-1)^{n-1}x^{n-1}$$ .

But I don't have a clue about how to find the minimum plynomial. is there is a way to find it without calculating the characteristic polynomial?

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In general, it is a good idea to calculate the characteristic polynomial, as the minimal polynomial is a divisor. In your case, note that $$A^k=n^{k-1}A.$$ So you can give the minimal polynomial immediately:

$A^2-nA = 0$ and it is clear that the degree is $\geq 2$, so $x^2-nx$ is the minimal polynomial.

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