TITLE: Cute problem: determinant of $I_n+(f_if_j)_{i,j}$
QUESTION [6 upvotes]: I thought of the following little problem.
Given numbers $f_1,\dots f_n$, what is the determinant of the symmetric matrix $I_n+(f_if_j)_{i,j}$?
I have found a cute combinatorial-style proof that it is $1+\Sigma_i f_i^2$. using the sum over permutations formula for the determinant. Here $F(\sigma)$ denotes the set of fixed points of $\sigma$.

Does anyone have a faster/more elegant method?

REPLY [2 votes]: Overkill but Sylvester's Theorem tells us:
$$\det(I_n + XY) = \det(I_m+YX)$$
for $X, Y$ of sizes of $n\times m$ and $m\times n$ respectively. Then for $F = (f_1,f_2,\dots f_n):$
$$\det(I_n + F^TF) = \det(I_1+FF^T) = 1+\sum_{i=1}^n f_i^2$$