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I have a sequence of Hankel matrices $A_{n}$ (i.e. matrix whose entries on all anti-diagonals are the same), and its entries are either $1$ or $0$. I am trying to prove that their determinant are either $0$ or $\pm 1$).

$$ A_{2}=\left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) $$

$$A_{3}= \left(\begin{matrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}\right) $$

$$A_{4}= \left(\begin{matrix} 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{matrix}\right) $$

$$ A_{5}=\left(\begin{matrix} 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{matrix}\right)\\ \vdots $$

That is, the main anti-diagonals are all $0$, and everything above the main anti-diagonals are $1$'s. The next anti-diagonals below the main anti-diagonals are $1$'s, and the rests are $0$'s.

I would appreciate it if anyone can help with this example or providing some non-trivial theorems regarding the determinant of Hankel matrix (non-trivial theorems means theorems that is beyond undergraduate linear algebra).

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  • $\begingroup$ @Milten That the question is closed for lack of clarity, so I posted a new one with more details. In addition, I intend to ask two questions, this one is about this particular example, and the other one (math.stackexchange.com/questions/4955759/…) is about theoretical part. $\endgroup$
    – Apple
    Commented Aug 9 at 15:00
  • $\begingroup$ @Shika Thanks very much for your reply! I just added $A_{5}$, would it help in understanding the general pattern of the matrix? In addition, maybe I miss something very obvious, why there is always a permutation of the column that gives an upper triangular matrix? For example, how to permute $A_{3}$ into an upper-triangular matrix? $\endgroup$
    – Apple
    Commented Aug 9 at 15:23
  • $\begingroup$ Actually I can't read: ignore the permutation part. $\endgroup$
    – Shika
    Commented Aug 9 at 15:25
  • $\begingroup$ @user8675309 Thanks for your comment! First, the sequences in these two questions are different. It is possible that two similar sequences turns out to be very different (e.g. the integration of $1/x^5$ and $1/(x^5+1)$). Second, that question math.stackexchange.com/questions/4955759/…) is not about that particular example, and it is about the theorems. That sequence is just an example to show what type of matrices I am interested in. $\endgroup$
    – Apple
    Commented Aug 9 at 15:48
  • $\begingroup$ @Apple I apologise, I didn’t notice the difference between the questions or the matrices. I’ll delete my initial comments :-) $\endgroup$
    – Milten
    Commented Aug 9 at 16:25

1 Answer 1

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It is an nice question, I would love to read more about where it comes from.

Here is a proof of the result:

Expanding $|A_n|$ along the last column, (you have only one non-zero term), you almost get $|A_{n-1}|$, except from a sign $(-1)^{n+2}$ and the first row which is full of $1$'s. Then using linearity (to isolate the last $1$), you get the the recurrence relation: $$|A_n|=(-1)^n|A_{n-1}|+|A_{n-2}|$$.

From $|A_2|=1$ and $|A_3|=-1$, you get successively: $$ 1,-1,0,-1,-1,0,-1,1,0,1,1,0,\underline{1,-1}\dots$$

The sequence is then periodic of order 12 and all its values are in $\{-1,0,1\}$.

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