Determinants of matrices defined by the minimum/maximum indices of their entries

Given $n$, let $L_n=\|l_{ij}\|$, $L’_n=\|l’_{ij}\|$, and $U_n=\|u_{ij}\|$, be $n\times n$ matrices such that $l_{ij}$ equals $1$, if $i\ge j$, and equals $0$, otherwise; $l’_{ij}$ equals $-1$, if $n-1\ge i\ge j$, equals $n$, if $i=n$, and equals $0$, otherwise; $u_{ij}$ equals $1$, if $j\ge i$, and equals $0$, otherwise. It is easy to check that $A_n=L_nU_n$ and $B_n=U_nL’_n$. Since $L_n$, $L’_n$, and $U_n$ are triangular matrices with $1$’s on the main diagonal, $\det L_n=\det U_n=1$ and $\det L’_n=(-1)^{n-1}n$, so $\det A_n=1$ and $\det B_n=(-1)^{n-1}n$. Alternatively, we can observe that $A_n J_n=-L_n$ and $J_nB_n=-L’_n$, where $J_n=-U_n^{-1}$ is an $n\times n$ Jordan cell with the eigenvalue $-1$. Since $\det J_n=(-1)^n$, we obtain the same conclusions.

Let us show the decompositions $A_n=L_nU_n$, $B_n=U_nL_n’$, $A_nJ_n=-L_n$, and $J_nB_n=-L’_n$ for $n=4$: $$ \begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2\\ 1 & 2 & 3 & 3\\ 1 & 2 & 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 \end{pmatrix}, $$

$$ \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 2 & 3 & 4\\ 3 & 3 & 3 & 4\\ 4 & 4 & 4 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0\\ -1 & -1 & 0 & 0\\ -1 & -1 & -1 & 0\\ 4 & 4 & 4 & 4 \end{pmatrix}, $$

$$\begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2\\ 1 & 2 & 3 & 3\\ 1 & 2 & 3 & 4 \end{pmatrix} \begin{pmatrix} -1 & 1 & 0 & 0\\ 0 & -1 & 1 & 0\\ 0 & 0 & -1 & 1\\ 0 & 0 & 0 & -1 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 & 0\\ -1 & -1 & 0 & 0\\ -1 & -1 & -1 & 0\\ -1 & -1 & -1 & -1 \end{pmatrix},$$ $$\begin{pmatrix} -1 & 1 & 0 & 0\\ 0 & -1 & 1 & 0\\ 0 & 0 & -1 & 1\\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 2 & 3 & 4\\ 3 & 3 & 3 & 4\\ 4 & 4 & 4 & 4 \end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ -4 & -4 & -4 & -4 \end{pmatrix}.$$

$ \newcommand{\RR}{\mathcal{R}} \newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)} \newcommand{\mdet}[1]{\left| \begin{matrix} #1 \end{matrix} \right|} $ Finding $\det(A_n)$:

I will pursue this one in part by simple row reduction: adding multiples of one row to another will preserve the determinant. So, for now, note the form of $A_n$:

$$A_n = \m{ 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 2 & \cdots & 2 \\ 1 & 2 & 3 & \cdots & 3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 2 & 3 & \cdots & n}$$

Throughout the entirety of this answer, let $\RR_i$ denote the $i$th row of the matrix, at whatever stage in the row reduction process we're at. We'll let $\sim$ denote equivalence by row reduction. At this point, subtract $\RR_1$ from each of $\RR_k$ for $k \ge 2$. We see that

$$A_n \sim \m{ 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 2 & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & \cdots & n-1}$$

Next, subtract $\RR_2$ from $\RR_1$. Then

$$A_n \sim \m{ 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 2 & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & \cdots & n-1}$$

Expanding the determinant along the first row (or column) trivially yields that

$$\det(A_n) = 1 \cdot \mdet{ 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 2 & \cdots & 2 \\ 1 & 2 & 3 & \cdots & 3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 2 & 3 & \cdots & n-1} = \det(A_{n-1})$$

This readily defines a linear homogenous recurrence relation, with initial condition $\det(A_1) = 1$, which sufficiently verifies that $\det(A_n) = 1$ for all $n \in \Bbb Z^+$. (This is particularly clear on back-iteration, which shows $\det(A_n) = \det(A_k)$ for $k \in \{1,2,\cdots,n-1\}$.)

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Finding $\det(B_n)$:

I will similarly utilize row reduction for this, as I did for $A_n$. In general, $B_n$ looks like this:

$$B_n = \m{ 1 & 2 & 3 & \cdots & n \\ 2 & 2 & 3 & \cdots & n \\ 3 & 3 & 3 & \cdots & n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n & n & n & \cdots & n}$$

Subtract $\RR_2$ from $\RR_1$ and we obtain

$$B_n \sim \m{ -1 & 0 & 0 & \cdots & 0 \\ 2 & 2 & 3 & \cdots & n \\ 3 & 3 & 3 & \cdots & n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n & n & n & \cdots & n}$$

Next, add $k \RR_1$ to each $\RR_k$ for $k \ge 2$:

$$B_n \sim \m{ -1 & 0 & 0 & \cdots & 0 \\ 0 & 2 & 3 & \cdots & n \\ 0 & 3 & 3 & \cdots & n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & n & n & \cdots & n}$$

Similarly, subtract $\RR_3$ from $\RR_2$, and then add $k \RR_2$ to each $\RR_k$ for $k \ge 3$.

$$B_n \sim \m{ -1 & 0 & 0 & \cdots & 0 \\ 0 & -1 & 0 & \cdots &0 \\ 0 & 0 & 3 & \cdots & n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & n & \cdots & n}$$

We proceed inductively. At the $r$th step, we subtract $\RR_{r+1}$ from $\RR_r$, turning $\RR_r$ into a row vector with all zeroes except $-1$ in its $r$th position. Then we add $k$ times this new $\RR_r$ to each $\RR_i$ with $i > k$, to negate their $r$th column and turn it into a similar row vector.

At the end of this process, after the $(n-1)$th step, we have that

$$B_n \sim \m{ -1 & 0 & 0 & \cdots & 0 \\ 0 & -1 & 0 & \cdots & 0 \\ 0 & 0 & -1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & n} = \mathrm{diag}(\underbrace{-1,-1,-1,\cdots,-1,-1}_{n-1 \; (-1)'s},n)$$

The determinant of $B_n$ is that of this diagonal matrix, and the determinant of a diagonal matrix is the product of the entries on the diagonal. Thus,

$$\det(B_n) = (-1)^{n-1} n$$

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Alternate Method for $\det(B_n)$:

This still utilizes row reduction, but more in line with how Dr. Peyam did it for $B_5$ in his video. So we instead subtract $\frac{n}{n-1} \RR_{n-1}$ from $\RR_n$. Then

\begin{align*} B_n &= \m{ 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 2 & 3 & \cdots & n-1 & n \\ 3 & 3 & 3 & \cdots & n-1 & n \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & n-1 & n-1 & \cdots & n-1 & n \\ n & n & n & \cdots & n & n} \\ &\sim \m{ 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 2 & 3 & \cdots & n-1 & n \\ 3 & 3 & 3 & \cdots & n-1 & n \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & n-1 & n-1 & \cdots & n-1 & n \\ 0 & 0 & 0 & \cdots & 0 & n - \frac{n^2}{n-1} } \end{align*}

We expand the determinant along the bottom row. Since the only nonzero entry is the bottom-right entry (i.e. $b_{n,n}^{(n)})$), the sign associated with this expansion is simply $+1$. (The sign of this in the cofactor expansion is more formally $(-1)^{n+n}=(-1)^{2n} = 1$, since the exponent is even.)

Thus,

\begin{align*} \det{B_n} &= \left( n - \frac{n^2}{n-1} \right) \mdet{ 1 & 2 & 3 & \cdots & n-1 \\ 2 & 2 & 3 & \cdots & n-1 \\ 3 & 3 & 3 & \cdots & n-1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n-1 & n-1 & n-1 & \cdots & n-1 } \\ &= \left( n - \frac{n^2}{n-1} \right) \det(B_{n-1}) \end{align*}

Visibly, one may proceed inductively, iterating backwards to the initial condition of $\det(B_1) = 1$, and observe

$$\det(B_n) = \prod_{k=2}^{n} \left( k - \frac{k^2}{k-1} \right)$$

One now notes that

$$ k - \frac{k^2}{k-1} = \frac{k(k-1) - k^2}{k-1} = \frac{k^2 - k - k^2}{k-1} = -\frac{k}{k-1}$$

and thus

$$\det(B_n) = \prod_{k=2}^{n} \left( -\frac{k}{k-1} \right)$$

This is a telescoping product, insofar as consecutive terms cancel nicely. Writing out the product, we have

\begin{align*} \require{cancel} \prod_{k=2}^{n} \frac{k}{k-1} &= \frac 2 1 \cdot \frac 3 2 \cdot \frac 4 3 \cdot \frac 5 4 \cdot \cdots \cdot \frac{n-2}{n-3} \cdot \frac{n-1}{n-2} \cdot \frac{n}{n-1}\\ &= \frac{\cancel{2}} 1 \cdot \frac {\cancel{3}} {\cancel{2}} \cdot \frac {\cancel{4}} {\cancel{3}} \cdot \frac {\cancel{5}} {\cancel{4}} \cdot \cdots \cdot \frac{\cancel {n-2}}{\cancel {n-3}} \cdot \frac{\cancel{n-1}}{\cancel {n-2}} \cdot \frac{n}{\cancel{n-1}}\\ &= n \end{align*}

Meanwhile, what remains now is the sign. Since it is constant with respect to $k$, we can pull it out of the product altogether. Note that the product has $n-1$ members. Thus,

$$\det(B_n) = (-1)^{n-1} \prod_{k=2}^{n} \frac{k}{k-1} = (-1)^{n-1} n$$

as desired.

Using the Schur complement,

$$\det \left( {\rm A}_{n+1} \right) = \det \begin{bmatrix} {\rm A}_n & {\rm A}_n \, {\rm e}_n\\ {\rm e}_n^\top {\rm A}_n & n+1 \end{bmatrix} = \underbrace{\left( n + 1 - {\rm e}_n^\top {\rm A}_n {\rm A}_n^{-1} {\rm A}_n \, {\rm e}_n \right)}_{= n + 1 - n} \det \left( {\rm A}_n \right) = \det \left( {\rm A}_n \right)$$

and, since $\det \left( {\rm A}_1 \right) = 1$, we conclude that $\det \left( {\rm A}_n \right) = 1$ for all $n \geq 1$.