Minimal polynomial of a matrix having only 1s on the counter diagonal 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.
 A: 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 .
A: 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.
A: 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 $A^2 = I$.
