Binomial determinant and LU decomposition

Question

Let $A_n$ be following $n \times n$ symmetric pentadiagonal matrix $$ \begin{pmatrix} 6&4&1&& \\ 4&\ddots&\ddots&\ddots&\\ 1&\ddots&\ddots&\ddots&1\\ &\ddots&\ddots&\ddots&4\\ &&1&4&6\\ \end{pmatrix}_{n\times n} $$ with entries of binomial expansion $(x+1)^4$.
I want to prove formula for its determinant $$ |A_n|=\frac{(n+1)(n+2)^2(n+3)}{12}. $$ There are several possibilities how to do it, for instance by solving recurrence relation from https://www.sciencedirect.com/science/article/pii/S0898122116300232 or by using method in https://www.cip.ifi.lmu.de/~grinberg/hyperfactorialBRIEF.pdf, but I want to prove it using LU decomposition just like as special case in paper https://www.ams.org/journals/mcom/1972-26-118/S0025-5718-1972-0303703-3/S0025-5718-1972-0303703-3.pdf#page16, but I don’t understand argumentation in proof of Theorem 1 - I don’t know why matrix $L_n$ is always lower triangular.
They are working with symmetric pentadiagonal matrix with entries of binomial expansion $(x-1)^4$, but their method could be easily transformed into my case by “cleverly putting” minus sign to matrix $U_n$. It seems to work, for example $n=6$: $$ A_6= \begin{pmatrix} 6&4&1&0&0&0\\ 4&6&4&1&0&0\\ 1&4&6&4&1&0\\ 0&1&4&6&4&1\\ 0&0&1&4&6&4\\ 0&0&0&1&4&6\\ \end{pmatrix} $$ $$ U_6= \begin{pmatrix} -1&-2&-3&-4&-5&-6\\ 0&3&6&9&12&15\\ 0&0&-6&-12&-18&-24\\ 0&0&0&10&20&30\\ 0&0&0&0&-15&-30\\ 0&0&0&0&0&21\\ \end{pmatrix} $$ and finally: $$ A_6 U_6= \begin{pmatrix} -6&0&0&0&0&0\\ -4&10&0&0&0&0\\ -1&10&-15&0&0&0\\ 0&3&-18&21&0&0\\ 0&0&-6&28&-28&0\\ 0&0&0&10&-40&36\\ \end{pmatrix}, $$ which is clearly lower triangular matrix and lets call it $L_6$.
The rest is easy using multiplicativity of determinant and fact determinant of lower/upper triangular matrix equals product of terms on the main diagonal: $$ |U_6|=(-1)(3)(-6)(10)(-15)(21) $$ $$ |L_6|=(-6)(10)(-15)(21)(-28)(36) $$ $$ |L_6|=|A_6 U_6|=|A_6||U_6|, $$ so $$ |A_6|=\frac{|L_6|}{|U_6|}=\frac{(-28)(36)}{(-1)(3)}=(28)(12)=336 $$ which equals closed form for $n=6$: $$ \frac{(6+1)(6+2)^2(6+3)}{12}=336. $$ My question is how to prove that for general $n$ using this method based on LU decomposition, or more specifically how to prove matrix $A_nU_n:=L_n$ is lower triangular for any $n$ (because the rest should be easy).

Answer 1

The paper which you have linked has

$$\begin{align} a_{ij} = (-1)^{r+i-j}\binom{2r}{r+i-j} & & U_{ij} &= \binom{i+r-1}{r}\binom{j+r-1-i}{r-1} \cdot [j \geq i] \end{align}$$

From this, it follows that for $L = AU$, we get

$$ L_{ij} = \sum\limits_k a_{ik} U_{kj} = \sum\limits_k (-1)^{r+i-k}\binom{2r}{r+i-k} \binom{k+r-1}{r} \binom{j+r-1-k}{r-1} \cdot [j \geq k] $$

In your case, however,

$$\begin{align} a_{ij}' = \binom{2r}{r+i-j} & & U_{ij}' &= (-1)^{i}\binom{i+r-1}{r}\binom{j+r-1-i}{r-1} \cdot [j \geq i] \end{align}$$

With this, the value of $L_{ij}'$ for $L' = A'U'$ can be found as

$$ L_{ij}' = \sum\limits_k a_{ik}' U_{kj}' = \sum\limits_k (-1)^k \binom{2r}{r+i-k} \binom{k+r-1}{r} \binom{j+r-1-k}{r-1} \cdot [j \geq k] = (-1)^{r+i} L_{ij} $$

Note that you have $r=2$, so basically the $i$-th row of $L'$ is simply the $i$-th row of $L$ multiplied by $(-1)^i$.

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Now, assuming that we do not know that $L$ from the article is lower-triangular, how to prove it?

The rationale from the article is that multiplication of $A$ by a column-vector $x(i)$ applies the transform

$$x(i) \to x(i+1)-2x(i)+x(i-1),$$

to it $r$ times in a row. This transform is called the second "central difference", and you can prove by induction that the coefficients after applying it $r$ times are the coefficients of $(x^2-2x+1)^r = (x-1)^{2r}$, which are, in turn, the coefficients of $A$.

You can also represent second central difference as the combination of transforms $x(i) \to x(i+1)-x(i)$ and $x(i) \to x(i) - x(i-1)$. From this, it follows that if $x(i)$ is a polynomial, applying second central difference to it $r$ times will reduce the degree of the polynomial by $2r$. Also note that the final value of $x(i)$ only depends on the initial values of $x(i-r),\dots,x(i+r)$, and not the elements beyond.

So, the article defines $U_{ij}$ in such way that for any $j$, when $i < j + r$, it holds that $U_{ij} = u_j(i)$, where

$$ u_j(i) = \binom{i+r-1}{r} \binom{j+r-1-i}{r-1}. $$

It means that first $j+r-1$ rows of the $j$-th column are the values of the same polynomial $u_j(i)$ in $i$ of degree $2r-1$, so after applying the second central difference to its values $r$ times, they will turn into $0$ for all $i$ such that $i+r<j+r$.

On the other hand, multiplying $A$ by $U$ is same as individually applying $A$ to each column of $U$. Therefore, in the $j$-th column of the result, first $j-1$ rows will be zero, proving that it would be a lower-triangular matrix. $\square$