let $x,y \in \mathbb C^n$ find the characteristic polynomial of $I_n +xy^*$.
I think we can use the relation $X^nP_{AB}(x)=X^mP_{BA}(X)$
we put $A=x$ and $B=x^{-1}+y^*$ is this right?
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gebafe
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What is $m$ here? – Dietrich Burde Feb 15 '23 at 17:17
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columns of matrix that is for all $M_{n,m}(\mathbb C)$ – gebafe Feb 15 '23 at 17:20
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For Googling purposes, such a matrix will be a "rank-one update to the identity". More generally there's a number of formulas for how various quantities change when a matrix is subject to a rank-one perturbation, for instance the matrix determinant lemma. – Semiclassical Feb 15 '23 at 17:21
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1Since x is a vector, x^{-1} is... what? – Mariano Suárez-Álvarez Feb 15 '23 at 17:23
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@MarianoSuárez-Álvarez $1/x$ – gebafe Feb 15 '23 at 17:25
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And that is what, exactly? – Mariano Suárez-Álvarez Feb 15 '23 at 17:25
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@MarianoSuárez-Álvarez it's $(1/x_1,1/x_2...)$ – gebafe Feb 15 '23 at 17:28
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And you are saying that AB is the same as I+x^{-1}y ? – Mariano Suárez-Álvarez Feb 15 '23 at 17:29
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No we have $AB= x(x^{-1}+y^)= 1+xy^=I_n+xy^*$ – gebafe Feb 15 '23 at 17:31
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Have you actually computed that in an example — with n=2, say? – Mariano Suárez-Álvarez Feb 15 '23 at 17:32
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Actually NO But am i wrong ? – gebafe Feb 15 '23 at 17:37
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@MarianoSuárez-Álvarez We can replace $I_n$ with $ee^\top$ ? – gebafe Feb 15 '23 at 17:49