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Consider the matrix $A=a_{ij}$ where $$a_{ij}=\begin{cases}1\ \ \text{if}\ \ i+j=n+1\\0\ \ \text{otherwise}\end{cases}$$. Then, what can be said about the minimal polynomial of the matrix $A$.

Note that one eigenvalue is easily found by taking the eigenvector $\begin{pmatrix}1\\1\\1\\\ldots\\\ldots\\\ldots\\1\end{pmatrix}$. Any hints. Thanks beforehand.

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    $\begingroup$ When $n>1$, what are $A-I,\ A+I$ and $(A-I)(A+I)$? $\endgroup$ – user1551 Jan 9 at 11:50
  • $\begingroup$ @user1551 how do we compute $(A-I)(A+I)$ for large matrices? Is there any inductive/recursive process? $\endgroup$ – vidyarthi Jan 9 at 11:53
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    $\begingroup$ Hint : We have $A^2=I$ , so the minimal polynomial must divide $x^2-1$ $\endgroup$ – Peter Jan 9 at 12:04
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Hint: Assuming $A$ is supposed to be an $n \times n$ matrix, $A^2$ is very easy to calculate, and this gives you your answer almost immediately. The entry in row $i$ and column $j$ of $A^2$ is the product of the $i^\mbox{th}$ row and $j^\mbox{th}$ colmn of $A$ , which you know.

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Thanks to @user1551, the answer is deceptively simple in this case. We have, $$A-I=\begin{pmatrix}-1&0&\ldots&0&1\\0&-1&\ldots&1&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\1&0&\ldots&0&-1\end{pmatrix}$$. Similarly, $$A+I=\begin{pmatrix}1&0&\ldots&0&1\\0&1&\ldots&1&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\1&0&\ldots&0&1\end{pmatrix}$$. Now multiplying would naturally give us the zero matrix as the $1$s and $-1$s cancel out in pairs at each non zero product entry .

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  • $\begingroup$ You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference). $\endgroup$ – user376343 Jan 9 at 19:25
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First by some computation, we can quickly find that its characteristic polynomial is

$ch_A(x) = (x-1)^{\big{\lceil}\frac{n}{2}\big{\rceil}}(x+1)^{\big{\lfloor}\frac{n}{2}\big{\rfloor}}$,

which means that $1$ and $-1$ are the only eigenvalues of $A$.

It turns out that geometric multiplicity of both eigenvalues are only one (i.e. $m(x) = x^2-1$), since $A^2 = I$.

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