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This is my first question, please forgive me if I mistake something, since I don't think I will be allowed to edit the question later.

So, let me explain the kind of matrices I'm talking about. Think of a $n\times n$ square matrix, and think of a sequence of $n^2$ consecutive squares starting at $k^2$ (the most natural choice being $k=1$). Now place those squares in the matrix as if you were writing, row by row. I'll dare to invent a notation for these "Square Matrice filled with Consecutive Squares": $SMCS(n,k)$. So we have, for example: $$ SMCS(3,4)= \begin{bmatrix} 4^2 & 5^2 & 6^2 \\ 7^2 & 8^2 & 9^2 \\ 10^2 & 11^2 & 12^2 \\ \end{bmatrix}= \begin{bmatrix} 16 & 25 & 36 \\ 49 & 64 & 81 \\ 100 & 121 & 144 \\ \end{bmatrix} $$ I wasn't able to find anything about matrices like these neither here nor elsewhere. Probably there are of no mathematical interest. My interest starts from a lecture, years ago, when the professor made us notice that the "Square Matrice filled with Consecutive Integers" $1$ to $9$ —it is the cell phone keypad! I'll call it $SMCI(3,1)$— is singular. That is: $$ \det \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{bmatrix}=0 $$ I find very intuitive that this elegant property holds for matrices of higher order and even if starting from a different integer, that is $$ \det \left[SMCI(n,k)\right]=0 \qquad n\geqslant 3, \;\forall k $$ I can recognize a clear pattern (where the middle culumns are an "average" of the ones at their sides) so that I feel relieved from the need to provide an explicit proof for this result, that shouldn't be too hard anyway.

Now I'm looking at the determinants of the matrices filled with squares. With no surprise I find nonsingular matrices of order $2$. Then, with some hope, I look at $SMCS(3,1)$ (the cell phone keypad with each number squared) but I find that the determinant is $-216$. So I find myself admitting that, understandably, the property doesn't hold for matrices filled with powers. Yet something curious happens: the result is the same even if I take a different starting point, that is $$ \det\left[SMCS(3,k)\right]=-216\qquad \forall k $$ This was a surprise, but it was pretty straightforward proving it with some direct algebra calculations.

Here comes the big deal. I went on looking at higher orders, wondering if there were a characteristic constant for each $n$. Instead... $0$'s began to appear again! Me unbeliever! I've grown convinced that $$ \det \left[SMCS(n,k)\right]=0 \qquad n\geqslant 4, \;\forall k $$ but attempting a direct proof, if only for $n=4$, was out of the question. Let alone a general proof for any $n$, which is what I would really be interested in, but I wouldn't know even where to start from. I don't really need the rigorous proof, what I'd love is to be sure of the validity of the property, and possibly a way to "understand" it in a manner similar to what I could do with the $SMCI$ matrices, to be "convinced" of the result. Can anyone help my with this? Thank you!


Wow, great, I got it! It took me some effort, but I got it all, thank you JeanMarie! Forgive me for coming back with some delay, but I saw you constantly improving your answer, and considering my time zone disadvantage I had to give up for the day.

Understanding your "trick" allows to easily extend the property to even higher orders; for example rows (or columns) of consecutive cubes give singular matrices from $5 \times 5$ on. If I extend the notation to "Square Matrix filled with Consecutive Powers", I shall write

$$ \det \left[SMCP(p,n,k)\right]=0 \qquad n\geqslant p+2, \;\forall k $$

I couldn't resist exploring the case just before the first singularity occurence, when the matrix order is greater than the power only by $1$. I mean, in the general case, because it is easy to algebraically verify that

$$ \det \left[SMCP(1,2,k)\right]=-2 \qquad \forall k $$

$$ \det \left[SMCP(2,3,k)\right]=-216 \qquad \forall k $$

I made a few direct attempts to find that (probably $\forall k$, but I don't dare writing it)

$$ \det \left[SMCP(3,4,k)\right]=5308416 $$

$$ \det \left[SMCP(4,5,k)\right]=7776 \cdot 10^{10} $$

I hoped to infer a rule from this sequence start, and try a proof at a later stage, but I cannot see anything. Nor I think that the kernel trick can be helpful at this. So the question now would be

$$ \det \left[SMCP(p,n,k)\right]=\;? \qquad \forall k \;\textrm{ when }\; n=p+1 $$

but I don't know how hard it can be to annswer it, and I won't ask anyone an eccessive effort about something that now has gone well beyond what I was looking for in the first place. I leave it here just in case someone is able to easily see something that doesn't occur to me. Again thank you!

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    $\begingroup$ First of all, I applaud your appetite for discovery and not giving up for lower dimensions. We need these kind of people in the mathematics community. $\endgroup$ – астон вілла олоф мэллбэрг Mar 30 '17 at 23:21
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    $\begingroup$ Besides, I appreciate your unusually elegant English. $\endgroup$ – Jean Marie Mar 31 '17 at 0:22
  • $\begingroup$ Thank you both! I don't believe I deserve so much consideration, nevertheless it makes me feel boosted! $\endgroup$ – lesath82 Mar 31 '17 at 9:14
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For orders $n > 3$, these matrices have a zero determinant because they always have a non zero vector in their kernel. Moreover, this vector is the same for all matrices with the same order:

$$\tag{1} \ \ \ \begin{pmatrix}\ \ 1\\-3\\ \ \ 3\\-1\end{pmatrix}, \ \ \ \ \begin{pmatrix}\ \ 1\\-4\\ \ \ 6\\-4 \\ \ \ 1\end{pmatrix}, \ \ \text{etc.}$$

depending if $n=4, 5, \cdots$ etc.

(more generally, a vector whose entries are the coefficients of $(1-x)^{n-1}$.)

This is not difficult to prove; for example in the $4 \times 4$ case, it is based on the identity:

$$\tag{2}\forall n \in \mathbb{Z}, \ \ n^2-3(n+1)^2+3(n+2)^2-(n+3)^2=0$$

The general case being almost as simple.

As you are clearly interested by understanding what is "behind the curtain" there is a deeper reason than somewhat "accidental" algebraic identities like in $(2)$.

I had the idea of using these vectors because I was rather familiar with finite differences calculus, which is parallel, up to a certain point, with "continuous" (differential) calculus. Have a look for example at (http://mathworld.wolfram.com/FiniteDifference.html).

In short: as we work with second degree expressions, multiplying by "masks" with $(1, \ \ -3, \ \ 3, \ \ -1)$ or $(1, \ \ -4, \ \ 6, \ \ -4, \ \ 1)$ coefficients amounts to differentiate three times, four times, etc... a second degree expression, "naturally" giving a zero result.

Remark 1: The null determinant property can be extended to a larger class of matrices, i.e., matrices having consecutive squares entries in a same row, but with a possible disruption with the next row, for example:

$$\begin{pmatrix}\ \ 1^2&2^2&3^2&4^2\\ 7^2&8^2&9^2&10^2\\ 3^2&4^2&5^2&6^2\\11^1&12^2&13^2&14^2\end{pmatrix}$$

Remark 2: I think that, beyond the fact that the determinant is $0$, a more accurate characterization is through the rank of the matrix. let us understand it on the example of $SMCS(5,7)$.

Let us denote by $M,V_1,V_2$ and $V_3$ the following matrix and vectors:

$$M=\begin{pmatrix}\ 49 & 64 & 81 & 100 & 121\\ 144 & 169 & 196 & 225 & 256\\ 289 & 324 & 361 & 400 & 441\\ 484 & 529 & 576 & 625 & 676\\ 729 & 784 & 841 & 900 & 961\end{pmatrix} \ \ \text{and} \ \ \ V_1=\begin{pmatrix}\ \ 1\\-3\\ \ \ 3\\-1 \\ \ \ 0\end{pmatrix}, V_2=\begin{pmatrix} \ \ 0 \\ \ \ 1\\-3\\ \ \ 3\\-1\end{pmatrix}, V_3=\begin{pmatrix}\ \ 1\\-4\\ \ \ 6\\-4 \\ \ \ 1\end{pmatrix}$$

(note that $V_1$ and $V_2$ are made with the first vector of (1), with a n added $0$ at the and or at the beginning.)

$\det(M)=0$ but with a more important rank loss than expected: rank$(M)=3=5-2$. It means that the kernel has dimension 2. It is interesting to see that a basis of the kernel is constituted by $V_1$ and $V_2$, with $V_3=V_1-V_2$ which is easily explained (the mask of the fourth discrete differentiation operator is the difference of the masks of the third discrete diff. operator).

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    $\begingroup$ I added my reply as an edit to the original question. I tried to do it here, but the maximum characters limit was too... limiting! $\endgroup$ – lesath82 Mar 31 '17 at 9:07
  • $\begingroup$ I will have a look at your "addendum". I have myself written in my answer a "remark 2" that pinpoints the interest for such matrices to consider the rank as conveying more information than the determinant. $\endgroup$ – Jean Marie Mar 31 '17 at 11:31
  • $\begingroup$ With some delay I come back to reason about your remark. I find it in perfect agreement with the finite difference explanation: $4$-vectors can represent a third derivative, which applied to a second degree function gives zero. Even more so will do higher dimensioned vectors representing subsequent derivations, and the kernel gains one dimension for every derivation of an already null expression. $\endgroup$ – lesath82 May 14 '17 at 13:23
  • $\begingroup$ Did you have a chance to look at my additional question? The mask vector $(1,-2,1)$ applied to a $3\times 3$ $SMCS$ gives a vector with constant entries $(2,2,2)$ but I cannot see whether this smart stratagem can help showing that the determinant has always the same value regardless of the starting point (in a general way valid for higher degrees/dimensions too). Do you confirm that this is probably no easy task? @JeanMarie $\endgroup$ – lesath82 May 14 '17 at 13:24
  • $\begingroup$ @lesath82 I will answer you (I am in a hurry right now). $\endgroup$ – Jean Marie May 14 '17 at 23:00

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