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A matrix

I was trying to calculate the determinant for every $n$ starting with $2$.
If $n$ is $2$ then determinant is $-1$, if $3$ is $-1$, if $4$ is $1$, if $5$ is $1$, if $6$ is $-1$, then I stopped.
It changes every $2 n$ but I don't know how to generalize it. Any idea?

EDIT: I need to use induction on $n$ but I don't know how.
Do I have to guess a formula first and then prove it with induction? If so, what's the formula?

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    Induction will do. – 2017-02-15
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    The '...' are poorly placed in that matrix -- are they intended to only cover 0's, other than the diagonal? – 2017-02-15
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    Looks like 1's only on the southwest to northeast diagonal, else 0's. – 2017-02-15
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    Yes it's 1's on the antidiagonal only – 2017-02-15
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    Induction? Use https://en.wikipedia.org/wiki/Laplace_expansion – 2017-02-15
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    That's what I'm trying to use. What I'm struggling about is that I don't have to prove a formula with induction, but I have to find a formula that proves induction. – 2017-02-15

3 Answers 3

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I found the solution!

Let's expand the determinant along the n-th row. We get:
$det(A_n)=(-1)^{n+1}det(A_{n-1})=(-1)^{n+1}(-1)^{n}...(-1)^{3}=$

since it's all $1$'s we only care about the sign
and since $n$ starts at $2$ the last power is $n+1=2+1=3$

$=(-1)^{(n+1)+n+...+3}=(-1)^{(1+2)+3+...+n+(n+1)-(1+2)}=(-1)^{\frac{(n+1)(n+2)}{2}-3}=(-1)^{\frac{n^2+3n-4}{2}}=(-1)^{\frac{(n-1)(n+4)}{2}}$

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Let $A$ be this matrix.

Swaping to columns changes only the sign of the déterminant.

You can figure out a sequence of columns swaps that start with this matrix and ends with the identity matrix (I will denote s(i,j) the operation that swaps columns $i$ and $j$) :

s(1,2) s(2,3) s(3,4) etc ... s(n-1,n) s(1,2) s(2,3) etc ... s(n-2,n-1) etc ... s(1,2)

Hence :

$$\det(A)=(-1)^{\frac{n(n-1)}{2}}$$

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The determinant does not change if you change the basis.

So if $e_{i}$ are the implicit basis vector here, rewrite the matrix with respect to the basis $$ e_{1} + e_{n}, e_{1} - e_{n}, e_{2} + e_{n-1}, e_{2} - e_{n-1}, \dots $$ where it may end with $$ e_{n/2} + e_{n/2+1} + , e_{n/2} - e_{n/2+1} $$ if $n$ is even, or with $$ e_{(n-1)/2} + e_{(n+3)/2} \qquad\text{and then}\qquad e_{(n+1)/2} $$ if $n$ is odd.

Note that on $e_{1} + e_{n}, e_{1} - e_{n}$ and similar blocks the matrix takes the form $$ \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}. $$ (For $n$ odd, $e_{(n+1)/2}$ is fixed by the matrix.) So the determinant is $(-1)^{n/2}$ if $n$ is even, and $(-1)^{(n-1)/2}$ if $n$ is odd. In short, it is $$ (-1)^{\left\lfloor \frac{n}{2} \right\rfloor}. $$

The answer is the same as the answer to this other question, because this matrix is the permutation matrix corresponding to the permutation there, and the determinant is the same as the sign of the permutation, by Leibniz's formula.

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    I'm having some trouble trying to decipher both answers. I can calculate a determinant and I may be able to use induction. The tools you used were not covered in my course. – 2017-02-15
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    @codew, sorry, Please just disregard my answer. – 2017-02-15