# 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.

• When $n>1$, what are $A-I,\ A+I$ and $(A-I)(A+I)$? – user1551 Jan 9 at 11:50
• @user1551 how do we compute $(A-I)(A+I)$ for large matrices? Is there any inductive/recursive process? – vidyarthi Jan 9 at 11:53
• Hint : We have $A^2=I$ , so the minimal polynomial must divide $x^2-1$ – Peter Jan 9 at 12:04

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.

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 .

• You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference). – user376343 Jan 9 at 19:25

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 $$A^2 = I$$.