# Roots of determinant of matrix with polynomial entries

Let $$p_1, p_2,\dots, p_n$$ and $$q_1,q_2,\dots,q_n$$ be a collection of complex polynomials. Let $$A$$ be a $$n \times n$$ matrix satisfying

$$a_{ij} = \begin{cases} p_i(x) & \text{ if } i = j, \\ q_i(x) & \text{ otherwise} \end{cases} .$$

is there any connection between the roots of the polynomials $$p_i$$'s and $$q_i$$'s and the roots of the polynomial $$\det A$$? if not, is this true under at least under any special assumptions?

Kindly share some references.

Thank you.

• If $r_i(x)=p_i(x)-q_i(x)$ then the determinant is $$\left(1+\sum_{i=1}^n \frac{q_i(x)}{r_i(x)}\right)\prod_{i=1}^n r_i(x).$$ Jul 16 '20 at 10:15
• @BrendanMcKay Thank you very much. Jul 16 '20 at 10:54

Let $$r_i := p_i - q_i$$.

$${\bf A} (x) := \begin{bmatrix} p_1 (x) & q_1 (x) & \ldots & q_1 (x)\\ q_2 (x) & p_2 (x) & \ldots & q_2 (x)\\ \vdots & \vdots & \ddots & \vdots\\ q_n (x) & q_n (x) & \ldots & p_n (x)\end{bmatrix} = \mbox{diag} \left( {\bf r} (x) \right) + {\bf q} (x) {\Bbb 1}_n^\top$$

Using the matrix determinant lemma,

$$\det \left ( {\bf A} (x) \right) = \det \left( \mbox{diag} \left( {\bf r} (x) \right) \right) \left( 1 + {\Bbb 1}_n^\top \mbox{diag}^{-1} \left( {\bf r} (x) \right) {\bf q} (x) \right) = \color{blue}{\left( 1 + \sum_{i=1}^n \frac{q_i(x)}{r_i(x)} \right)\displaystyle\prod_{i=1}^n r_i (x)}$$

as mentioned by Brendan McKay some 20 minutes ago.

• Thank you. The reference is very helpful. Jul 16 '20 at 10:55
• is it possible to do this, if I assume some entries if this matrix are zero other entries are as it is? Thank you. Jul 16 '20 at 13:39
• @GA316 The matrix determinant lemma is used for rank-$1$ updates of invertible matrices. I assume it can be generalized to non-invertible matrices. What kind of matrices do you have in mind? Jul 16 '20 at 13:57
• Consider your matrix $A(x)$. Now, I make some entries zero randomly such a way the if $a_{ij} =0$ then $a_{ji} = 0$. Remaining entries of $A(x)$ are unchanged. Thank you. Jul 16 '20 at 14:11
• @GA316 Once you destroy the "invertible matrix + rank-1 matrix" structure, you have to use another approach. Jul 16 '20 at 14:28