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?


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

  • 1
    $\begingroup$ Sylvester? Typo $\endgroup$ – Benjamin Wang Feb 2 at 18:56
  • $\begingroup$ @BenjaminWang yes $\endgroup$ – dezdichado Feb 3 at 14:06

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