Compute an $n \times n$ determinant

Question

Compute the following determinant

$$ \begin{vmatrix} a_{1}-b_{1}+x & a_{1}-b_{2} & \ldots & a_{1}-b_{n} \\ a_{2}-b_{1} & a_{2}-b_{2}+x & \ldots & a_{2}-b_{n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n}-b_{1} & a_{n}-b_{2} & \ldots & a_{n}-b_{n}+x \end{vmatrix}. $$

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I tried for $n=3$ and I used the fact that $\det$ funcion is liniar in each column.

\begin{align*} & \begin{vmatrix} a_{1}-b_{1}+x & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2}+x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x \end{vmatrix} \\ & = \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2}+x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x \end{vmatrix} +% \begin{vmatrix} x & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2}+x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x \end{vmatrix} \\ & =% \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x \end{vmatrix} +% \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x \end{vmatrix} \\ & +% \begin{vmatrix} x & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x \end{vmatrix} +% \begin{vmatrix} x & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x \end{vmatrix} \\ & =% \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}% \end{vmatrix} +% \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & x \end{vmatrix} \\ & +% \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}% \end{vmatrix} +% \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & x \end{vmatrix} \\ & +% \begin{vmatrix} x & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}% \end{vmatrix} +% \begin{vmatrix} x & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & x \end{vmatrix} \\ & +% \begin{vmatrix} x & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}% \end{vmatrix} +% \begin{vmatrix} x & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & x \end{vmatrix} \\ & =% \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}% \end{vmatrix} +% \begin{vmatrix} a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\ a_{2}-b_{1} & x & a_{2}-b_{3}\\ a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}% \end{vmatrix} \\ & +% \begin{vmatrix} x & 2\left( a_{1}-b_{2}\right) & a_{1}-b_{3}\\ a_{2}-b_{1} & a_{2}-b_{2}+x & a_{2}-b_{3}\\ a_{3}-b_{1} & 2\left( a_{3}-b_{2}\right) & a_{3}-b_{3}% \end{vmatrix} +% \begin{vmatrix} x+a_{1}-b_{1} & 2\left( a_{1}-b_{2}\right) & a_{1}-b_{3}\\ 2\left( a_{2}-b_{1}\right) & x+a_{2}-b_{2} & a_{2}-b_{3}\\ 2\left( a_{3}-b_{1}\right) & 2\left( a_{3}-b_{2}\right) & x \end{vmatrix} \end{align*} From here I didn't see any pattern or some manipulations to do in order to compute the determinant. The first determinants after the last $=$ sign is equal to zero after some calculation using Sarrus rule. How can one compute this determinant?

Answer 1

This equals $\det (xI+A-B)$ where $$A=\pmatrix{a_1&a_1&\cdots&a_1\\a_2&a_2&\cdots&a_2\\\vdots&\vdots&\ddots&\vdots\\a_n&a_n&\cdots&a_n}$$ and $$B=\pmatrix{b_1&b_2&\cdots&b_n\\b_1&b_2&\cdots&b_n\\\vdots&\vdots&\ddots&\vdots\\b_1&b_2&\cdots&b_n}.$$ Then $\det(xI+A-B)$ is the characteristic polynomial of $C=B-A$.

But both $A$ and $B$ have rank $\le 1$, so $C$ has rank $\le 2$. So all $d\times d$ minors of $C$ vanish for $d\ge3$ and so $$\det(xI-C)=x^{n}-t_1x^{n-1}+t_2x^{n-2}.$$ Here $t_1$ is the trace of $C$: $$t_1=\sum_i(b_i-a_i).$$ Also, $$t_2=\sum_{i<j}\left[(b_i-a_i)(b_j-a_j)-(b_i-a_j)(b_j-a_i)\right].$$

Answer 2

Given vectors $\rm{a}, \rm{b} \in \mathbb R^n$, where $n \geq 2$, let function $f : \mathbb R \to \mathbb R$ be defined by

$$f (x) := \det \left( x \rm{I}_n + \rm{a} \Bbb{1}_n^\top - \Bbb{1}_n \rm{b}^\top \right)$$

Using the matrix determinant lemma,

$$\begin{aligned} f (x) &= \det \left( x \rm{I}_n + \begin{bmatrix} | & | \\ \rm{a} & -\Bbb{1}_n\\ | & | \end{bmatrix} \begin{bmatrix} | & | \\ \Bbb{1}_n & \rm{b} \\ | & | \end{bmatrix}^\top \right) \\ &= \det \left( x \rm{I}_2 + \begin{bmatrix} | & | \\ \Bbb{1}_n & \rm{b} \\ | & | \end{bmatrix}^\top \begin{bmatrix} | & | \\ \rm{a} & -\Bbb{1}_n\\ | & | \end{bmatrix} \right) \cdot x^{n-2}\\ &= \det \left( x \rm{I}_2 + \begin{bmatrix} \Bbb{1}_n^\top \rm{a} & -n\\ \rm{a}^\top \rm{b} & -\Bbb{1}_n^\top \rm{b}\end{bmatrix} \right) \cdot x^{n-2}\\ &=\left( x^2 + \mbox{tr} \begin{bmatrix} \Bbb{1}_n^\top \rm{a} & -n\\ \rm{a}^\top \rm{b} & -\Bbb{1}_n^\top \rm{b}\end{bmatrix} x + \det \begin{bmatrix} \Bbb{1}_n^\top \rm{a} & -n\\ \rm{a}^\top \rm{b} & -\Bbb{1}_n^\top \rm{b}\end{bmatrix}\right) \cdot x^{n-2}\\ &= \color{blue}{x^n + \Bbb{1}_n^\top \left(\rm{a} - \rm{b} \right) x^{n-1} + \left( n \, \rm{a}^\top \rm{b} - \Bbb{1}_n^\top \rm{a} \rm{b}^\top \Bbb{1}_n \right) x^{n-2}} \end{aligned}$$