Let $A_{n}$ be the matrix with zeros on the diagonal and ones everywhere else. Show that the $\det(A_n) = 1-n $ .
This is the question that my professor asked but I think it must be a typo. Is it not $(-1)^n(1-n)$?
For the proof I was thinking of doing it by induction but I'm a little confused about the inductive step.
There is a useful result that $\det (I-u v^T) = 1-u^T v$.
If we let $e = (1,...,1)^T$ then $A_n = e e^T -I$, and so $\det A_n = (-1)^n \det (-A_n) = (-1)^n \det (I - e e^T) = (-1)^n (1-e^T e) = (-1)^n (1-n)$.
Addendum: To see the first result, note that for any $A,B$ with compatible dimensions, $AB$ and $BA$ have the same non zero eigenvalues. Then $I-AB$ and $I-BA$ have the same 'non one' eigenvalues. Since the determinant is the product of the eigenvalues, we have $\det (I-AB) = \det (I-BA)$. Choosing $A=u,B=v^T$ yields the desired result.
Yes, $\left(-1\right)^n\left(1-n\right)$ is correct. Here is a generalization distinct from @copper.hat's one:
Let $n\in\mathbb{N}$. Let $a_{1},a_{2},\ldots,a_{n}$ be $n$ elements of a commutative ring $\mathbb{K}$. Let $x\in\mathbb{K}$. Prove that $$ \det \begin{pmatrix} x & a_{1} & a_{2} & \cdots & a_{n-1} & a_{n}\\ a_{1} & x & a_{2} & \cdots & a_{n-1} & a_{n}\\ a_{1} & a_{2} & x & \cdots & a_{n-1} & a_{n}\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{1} & a_{2} & a_{3} & \cdots & x & a_{n}\\ a_{1} & a_{2} & a_{3} & \cdots & a_{n} & x \end{pmatrix} =\left( x+\sum_{i=1}^{n}a_{i}\right) \prod_{i=1}^{n}\left( x-a_{i}\right) . $$
(Exercise 6.21 in my Notes on the combinatorial foundations of algebra, as of 10 January 2019. No claims of originality made -- I've lifted this exercise from Yisong Yang's A Concise Text on Linear Algebra, exercise 3.2.8; only the solution is different. The exercise also appears as Exercise 258 in: D. Faddeev and I. Sominsky, Problems in Higher Algebra, Mir 1972.)
