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.

  • 5
    $\begingroup$ If $r_i(x)=p_i(x)-q_i(x)$ then the determinant is $$\left(1+\sum_{i=1}^n \frac{q_i(x)}{r_i(x)}\right)\prod_{i=1}^n r_i(x).$$ $\endgroup$ Jul 16 '20 at 10:15
  • $\begingroup$ @BrendanMcKay Thank you very much. $\endgroup$
    – GA316
    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.

  • $\begingroup$ Thank you. The reference is very helpful. $\endgroup$
    – GA316
    Jul 16 '20 at 10:55
  • $\begingroup$ is it possible to do this, if I assume some entries if this matrix are zero other entries are as it is? Thank you. $\endgroup$
    – GA316
    Jul 16 '20 at 13:39
  • $\begingroup$ @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? $\endgroup$ Jul 16 '20 at 13:57
  • $\begingroup$ 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. $\endgroup$
    – GA316
    Jul 16 '20 at 14:11
  • $\begingroup$ @GA316 Once you destroy the "invertible matrix + rank-1 matrix" structure, you have to use another approach. $\endgroup$ Jul 16 '20 at 14:28

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