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This question already has an answer here:

I have been trying to solve a problem and have given my answer in the form of a determinant, namely: \begin{align*} \det\begin{pmatrix} \binom{q}{2} & q & 1 & 0 & \dots & 0 \\ \binom{q}{3} & \binom{q}{2} & q & 1 & \dots & 0 \\ \binom{q}{4} & \binom{q}{3} & \binom{q}{2} & q & \dots & 0 \\ \binom{q}{5} & \binom{q}{4} & \binom{q}{3} & \binom{q}{2} & \dots & 0 \\ \vdots & \vdots & \vdots & \dots & \ddots & q \\ \binom{q}{n+1} & \binom{q}{n} & \binom{q}{n-1} & \binom{q}{n-2} & \dots & \binom{q}{2} \end{pmatrix} \end{align*} This is probably enough, but I thought I'd challange myself and dust off my old linear algebra skills. Unfortunately, it is not going well. I have attempted brute forcing it by multiplying out coefficients, but I don't think that is the way to go, because I am getting nowhere. Everytime I touch it I seem to be increasing the complexity rather than decreasing it...

Is there a good way to tackle this problem, or is this as good as it gets?

Any help would be much appreciated!

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marked as duplicate by darij grinberg, user1551 linear-algebra Jan 19 at 0:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is the particular case of math.stackexchange.com/questions/2681139 for $l=q$ and $k=2$. $\endgroup$ – darij grinberg Jan 18 at 23:49
  • $\begingroup$ I completed my answer to your question on universal coverings of $T^*$ about 3 minutes after you deleted it. If you would like to undelete it, I will post my answer. $\endgroup$ – Lee Mosher Apr 30 at 13:20
  • $\begingroup$ @LeeMosher my apologies! I never meant to delete it. It's up again $\endgroup$ – KurtKnödel Apr 30 at 14:02
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seems quite rigid pattern; here my number would be your $n+1.$ I put in the one with 7 so you could see the coefficient of the highest degree term. I think the constants factor in a similar way, as $$ 144 = 1 \cdot 2^2 \cdot 3^2 \cdot 4, $$ $$ 2880 = 1 \cdot 2^2 \cdot 3^2 \cdot 4^2 \cdot 5, $$ $$ 86400 = 1 \cdot 2^2 \cdot 3^2 \cdot 4^2 \cdot 5^2 \cdot 6, $$ $$ 3628800 = 1 \cdot 2^2 \cdot 3^2 \cdot 4^2 \cdot 5^2 \cdot 6^2 \cdot 7, $$

That leads to a solid recursion, let me call my number $w,$ $$ d_{w+1} = \left( \frac{(q+w-2)(q+w-3)}{w(w-1)} \right) \; d_w $$ See if you can write that securely in your notation.

One thing that comes out is a closed form. If the lower left element has $n+1,$ the determinant comes out $$ \frac{1}{n+1} \left( \begin{array}{c} q+n-1 \\ n \end{array} \right) \left( \begin{array}{c} q+n-2 \\ n \end{array} \right) $$

? det7 = matdet(d)
%26 = 1/3628800*q^12 + 1/151200*q^11 + 7/103680*q^10 + 23/60480*q^9 + 1541/1209600*q^8 + 1/400*q^7 + 1747/725760*q^6 - 13/60480*q^5 - 383/129600*q^4 - 101/37800*q^3 - 1/1260*q^2


===================================================================

? factor(det4 * 144)
%16 = 
[q - 1 1]

[    q 2]

[q + 1 2]

[q + 2 1]


? factor(det5 * 2880)
%15 = 
[q - 1 1]

[    q 2]

[q + 1 2]

[q + 2 2]

[q + 3 1]



? factor(det6 * 86400)
%20 = 
[q - 1 1]

[    q 2]

[q + 1 2]

[q + 2 2]

[q + 3 2]

[q + 4 1]

? 
? factor(det7 * 3628800)
%27 = 
[q - 1 1]

[    q 2]

[q + 1 2]

[q + 2 2]

[q + 3 2]

[q + 4 2]

[q + 5 1]

=========================================================================
? factor(144)
%21 = 
[2 4]

[3 2]

? factor(2880)
%22 = 
[2 6]

[3 2]

[5 1]

? factor(86400)
%23 = 
[2 7]

[3 3]

[5 2]

? 
? factor( 3628800)
%28 = 
[2 8]

[3 4]

[5 2]

[7 1]
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  • $\begingroup$ Thanks a lot, that is indeed a very nice recursion. I have to admit I am still having some difficulties proving the recursion. Is there any intuition behind it? $\endgroup$ – KurtKnödel Jan 17 at 23:02
  • $\begingroup$ @KurtKnödel Time will tell. It does become a closed form, I typed that in now. Same message, see if you can match that up with the way you are writing things. $\endgroup$ – Will Jagy Jan 17 at 23:33
  • $\begingroup$ One more cheeky question. My current approach is expanding the determinant by its first row. Do you think that this is the right approach, or am I way off? $\endgroup$ – KurtKnödel Jan 18 at 10:48
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    $\begingroup$ Update: I managed to prove it! Thanks a lot!:) $\endgroup$ – KurtKnödel Jan 18 at 14:18
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I cannot give an analytic proof, but this seems to give the correct results:

$$D=\frac{1}{n!(n+1)!}\frac{(q+n-1)!}{(q-1)!}\frac{(q+n-2)!}{(q-2)!}$$

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