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Suppose $u_k, v_k\in R^n$, $k=1, \ldots, m$. Define $F=\sum_{k=1}^m u_kv_k'$. How to show that $\det(I+F)=\det(\delta_{jk}+u_j'v_k)_{j,k=1}^m$? Here $I$ is the identity matrix, $u'$ is transpose of $u$, $\delta_{jk}$ is the Kronecker delta.

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up vote 2 down vote accepted

If $U$ is the matrix with columns $u_k$ and $V$ the matrix with columns $v_k$, then $F = UV'$ while $(u_j' v_k) = U'V$. The nonzero eigenvalues of $U'V$ and $UV' = (VU')'$, say $\lambda_1, \ldots, \lambda_r$ counted by multiplicity, are the same. However, $UV'$ is $n \times n$ while $V'U$ is $m \times m$. So the characteristic polynomials are $P_{U'V} = \det(\lambda I - U'V) = \lambda^{n-r} \prod_{i=1}^r (\lambda - \lambda_i)$ while $\det(\lambda I - UV') = \lambda^{m-r} \prod_{i=1}^r (\lambda - \lambda_i)$. In particular, take $\lambda = -1$ and we find $\det(I+U'V) = (-1)^n \det(-I-U'V) = (-1)^m \det(-I-UV') =\det(I+UV')$.

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Dear Robert Israel, It is my pleasure to have your comments on the following question:… – Blind Jul 3 '15 at 16:54

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