# Cute problem: determinant of $I_n+(f_if_j)_{i,j}$ [duplicate]

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?

The eigenvalues of $$(f_if_j)$$ are $$(\|f\|^2, 0,\dots,0)$$ so those of your matrix are $$(1+\|f\|^2, 1,\dots,1)$$
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 $$F = (f_1,f_2,\dots f_n):$$ $$\det(I_n + F^TF) = \det(I_1+FF^T) = 1+\sum_{i=1}^n f_i^2$$