0
$\begingroup$

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?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.