For each $n\ge2$ calculate the determinant of this matrix

Question

The matrix is

$\left(\begin{matrix} 0 & 0 & \cdots&0&1 \\ 0 & 0 & \cdots&1&0 \\ \vdots&\vdots&\ddots & \vdots & \vdots\\ 0&1&\cdots&0&0\\ 1&0&\cdots&0&0 \end{matrix}\right)$

For example for $n=2$ the matrix is

$\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right)$

For $n=3$

$\left(\begin{matrix} 0 &0& 1 \\ 0&1&0\\ 1 &0&0 \end{matrix}\right)$

So by making some calculations I noticed that the same value of the determinant is repeated at regular intervals. In particular $-1,-1,1,1,-1,-1... $ for $n=2,3,4,5,6,7...$. I have some difficulties in formalising this result and maybe answer with a closed formula (?) to determine what is the determinant for for whatever $n$ I choose. Any hint?

Answer 1

Notice that $A^2 = I_n$ where $I_n$ is the identity matrix of size $n$. Therefore, A is diagonizable and all eigenvalues of A are either $1$ or $-1$. Let $k$ be the multiplicity of the eigenvalue $1$ and $l$ the multiplicity of $-1$. We have $\det(A) = (-1)^l$ and $k +l =n$. Taking the trace of $A$, we have $k-l=0$ if $n$ is even, and $k-l=1$ if $n$ is odd. Therefore $\det(A) = (-1)^{n/2}$ if $n$ is even, and $\det(A)=(-1)^{(n-1)/2}$ otherwise. And it is always true that $\det(A) = (-1)^{n(n-1)/2}$

I hope this answers your question.

Answer 2

Let $d_n$ the desired determinant. Expand it along the first line you get

$$d_n=(-1)^{n+1}d_{n-1}=(-1)^{n-1}d_{n-1}$$ hence by induction $$d_n=(-1)^{\sum_{k=2}^{n-1}k}d_2=(-1)^{{\sum_{k=1}^{n-1}k}}=(-1)^{n(n-1)/2}$$

Answer 3

A basic property of determinants is that swapping two columns of the matrix reverses the sign of the determinant. Observe that you can obtain this matrix from the identity by swapping column $1$ with column $n$, column $2$ with column $n-1$, and so on. This means that the determinant of this matrix must be equal to $\pm1$, with the sign depending on how many swaps were required: If $n$ is even, then $n/2$ swaps are done; if $n$ is odd, it’s $(n-1)/2$, since the central column remains untouched. This can be written succinctly using the floor function: the determinant is equal to $(-1)^{\left\lfloor\frac n2\right\rfloor}$.

Floor and ceiling can be inconvenient to manipulate in expressions. We really only care about the parity of the number of swaps—whether it’s even or odd—so you could use any other simple function of $n$ that has the same parity as $\left\lfloor\frac n2\right\rfloor$ such as $n(n-1)/2$.