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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 $\sigma$.

enter image description here Does anyone have a faster/more elegant method?

math_lover
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    Some solutions here: https://math.stackexchange.com/questions/1101657/find-the-determinant-of-the-following-matrix and here: https://math.stackexchange.com/questions/730134/determinant-of-rank-one-perturbation-of-a-diagonal-matrix. See also https://math.stackexchange.com/q/420198/42969. – Martin R Feb 02 '21 at 18:57

2 Answers2

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The eigenvalues of $(f_if_j)$ are $(\|f\|^2, 0,\dots,0)$ so those of your matrix are $(1+\|f\|^2, 1,\dots,1)$

Bananach
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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 $F = (f_1,f_2,\dots f_n):$ $$\det(I_n + F^TF) = \det(I_1+FF^T) = 1+\sum_{i=1}^n f_i^2$$

dezdichado
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