Prove that a sequence of matrices have determinant $0$ or $\pm 1$.

Question

I have a sequence of matrix $A_{n}$, whose entries are either $1$ or $0$. I am trying to prove that they all have determinant are $0$ or $\pm 1$.

For example,

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

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

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

$$ A_{6}=\left(\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{matrix}\right)\\ \vdots $$

(this is a particular example, in general, they need not be symmetric or being a Hankel matrix.)

I would appreciate it if anyone can provide me with some advanced theorems/paper regarding the determinant of matrix (advanced theorems means theorems beyond undergraduate linear algebra).

@Qiaochu Yuan's answer is close to what I am looking for, and I would appreciate it if anyone can provide more.

Answer 1

Edit: Your example matrices appear to have nice LU decompositions, with diagonal entries $\pm 1$; if you could prove that in general you'd have your result. Also, performing one cofactor expansion appears to ruin their nice inductive structure but maybe performing two cofactor expansions could reveal some recursion.

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This is usually not how these calculations go (where you are trying to calculate the determinants of a sequence of matrices). Often you'll instead want to prove some kind of recurrence relation relating $\det A_n$ to $\det A_{n-1}$ or other previous terms, e.g. via cofactor expansion or something similar. Of course depending on the structure of the matrices something quite different might be needed; it really depends on the details of what the matrices look like.

Here is an example. Consider the sequence of matrices $M_n$ whose entries are given by the Catalan numbers in the form $(M_n)_{ij} = C_{i+j} = \frac{1}{i + j + 1} {2(i + j) \choose i + j}$, where $0 \le i, j \le n-1$; for example

$$M_3 = \begin{bmatrix} C_0 & C_1 & C_2 \\ C_1 & C_2 & C_3 \\ C_2 & C_3 & C_4 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 5 \\ 2 & 5 & 14 \end{bmatrix}.$$

Then it turns out that $\det M_n = 1$ for all $n$. This is really not obvious, and none of the statements in the comments help prove it! A proof can be given using some general facts about orthogonal polynomials, or using the Lindström-Gessel-Viennot lemma. The matrices $M_n$ also appear to have a nice Cholesky decomposition with a pattern that looks related to the Catalan triangle, and if you could prove that that would also work.

For a survey of some of the different ways complicated determinant evaluations can go you can check out Krattenthaler's Advanced Determinant Calculus and Advanced Determinant Calculus: A Complement.