factorization of the product of a matrix element and its cofactor

Question

Hi,

this is kind of continuation of this thread to concentrate on a specific problem from linear algebra and analysis that, I think, is rather interesting for itself. Here we go:

1) Main problem: Let $(c_n)_{n\in\mathbb{N}}$ be some fixed sequence of complex numbers. Is there a sequence of matrices $A_n=(a(n)_{ij})\in M_n(\mathbb{C})$, $n\in\mathbb{N}$, such that $a(n)_{ij}\widetilde{a(n)_{ij}}=c_i c_j$, where $\widetilde{a(n)_{ij}}$ denotes the cofactor of $a(n)_{ij}$? (For a fixed $n\in\mathbb{N}$ this actually establishes a non-linear system of $n^2$ equations with $n^2$ unknowns.)

2) "Inverse" problem: Let $A_n=(a(n)_{ij})\in M_n(\mathbb{C})$, $n\in\mathbb{N}$, be some given sequence of complex quadratic matrices of increasing order. What are necessary and sufficient conditions for such a matrix sequence to have a sequence of complex numbers $(c_n)_{n\in\mathbb{N}}$ such that $\forall n\in\mathbb{N}$ $\forall 1\leq i \leq n, 1\leq j \leq n: a(n)_{ij}\widetilde{a(n)_{ij}}=c_i c_j$ (notation as above)?

Any input is welcome and I highly appreciate any references to similar or related problems.

Thanks,

efq

Answer 1

For the main problem, we can use expansion by minors along the ith row to compute

$\det(A_n) = (-1)^{i+1}(a(n)_{i1}\widetilde{a(n)_{i1}} - a(n)_{i2}\widetilde{a(n)_{i2}} + a(n)_{i3}\widetilde{a(n)_{i3}} - \dots)$

or det(A_n) = (-1)ⁱ⁺¹c_i(c₁ - c₂ + c₃ - ...).

This is true for any i <= n, so whenever c₁ - c₂ + ... - (-1)ⁿc_n != 0 we must have c_i = (-1)ⁱ⁺¹c₁ for all i <= n. Therefore when c_n != (-1)ⁿ⁺¹c₁ for the first time, not only must we have c₁ - c₂ + ... - (-1)ⁿc_n = 0 but this alternating sum must vanish for all greater n, meaning that c_j = 0 for all j > n.

In conclusion: this is only possible if c_k = (-1)^k+1c₁ for all k < K (K constant but possibly infinite) and then c_k = 0 for all k > K.