0

$$ \begin{vmatrix} 1 + 2 a_1 & a_1 + a_2 & a_1+a_3 & \cdots & a_1+a_n\\ a_2 + a_1 & 1 + 2 a_2 & a_2+a_3 & \cdots & a_2+a_n\\ a_3 + a_1 & a_3 + a_2 & 1+2a_3 & \cdots & a_3+a_n\\ \vdots & \vdots & \vdots & \ddots &\vdots\\ a_n + a_1 & a_n + a_2 & a_n+a_3 & \cdots & 1+2a_n\\ \end{vmatrix} $$

Most likely there is a very simple solution here, but I just can’t see it. Please give me some hints. I tried subtracting the first row from all the others and then subtracting the last column from all the columns, it seemed like I was close, but it didn't get me anywhere.

Rodrigo de Azevedo
  • 1
Михаил Расторгуев
  • 11
  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Martin R Dec 19 '23 at 12:51
  • 3
    Hint: you can write it as identity plus a 2 rank symmetric matrix – Exodd Dec 19 '23 at 13:24
  • @Exodd I am new to Linear algebra and trying to figure out how this hint helps. $\det(A+I)$ where $A$ is symmetric and $\text{rank}(A)=2$. But I don't think $-1$ is an eigenvalue of $A$ so the answer is non-zero? – Nothing special Dec 19 '23 at 13:39

2 Answers2

2

Denote $b=(1,\ldots,1)^T$, $a=(a_1,\ldots,a_n)^T$, and $A=ab^T+ba^T.$ Assume for a while that $a$ and $b$ are independent. Consider the two dimensional space $F\subset R^n$ generated by $a$ and $b$ and compute $A(\alpha a+\beta b)$ in order to

  1. observe that $F$ is stable by $A$
  2. prove that the two eigenvalues of $A$ restricted to $F$ are $\langle a,b\rangle\pm\|a\|\|b\|.$

Finally $A$ is a symmetric matrix and any vector $v$ orthogonal to $F$ satisfies $Av=0.$ As a result the eigenvalues of $A$ are $\langle a,b\rangle\pm\|a\|\|b\|$ and $0$ with multiplicity $n-2.$ The eigenvalues of $I+A$ are $1+\langle a,b\rangle\pm\|a\|\|b\|$ and $1$ with multiplicity $n-2.$ The determinant of $I+A$ is $$(1+\langle a,b\rangle)^2-\|a\|^2\|b\|^2.$$ If $a$ and $b$ are dependent, the result is still true, by considering $a^k$ and $b^k$ independent such that $a^k\to_ka$ and $b^k\to _k b.$

Sorry for offering a standard solution for a student of second year, and not a clever one for a student of first year (that I have not really looked for).

Gérard Letac
  • 4,357
0

Let ${\bf a} := \begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}^\top$. Using Weinstein-Aronszajn,

$$ \det \left( {\bf I}_n + {\bf a} {\bf 1}_n^\top + {\bf 1}_n {\bf a}^\top \right) = \cdots = 1 - n \, \underbrace{{\bf a}^\top \left( {\bf I}_n - \frac{{\bf 1}_n {\bf 1}_n^\top}{n} \right) {\bf a}}_{= \operatorname{Var} ({\bf a})} + 2 \,{\bf 1}_n^\top {\bf a} $$

where $\operatorname{Var} ({\bf a})$ denotes the variance of $\bf a$.

Rodrigo de Azevedo
  • 1