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The matrix is \begin{equation*} M = \begin{pmatrix} a_1b_1+1 & a_1b_2 & \cdots & a_1b_n \\ a_2b_1 & a_2b_2+1 & \cdots & a_2b_n \\ \vdots & \vdots & \ddots & \vdots \\ a_nb_1 & a_{m,2} & \cdots & a_nb_n+1 \end{pmatrix} \end{equation*}

This matrix is decomposable to be the sum of identity matrix plus $\alpha^T \beta$, where $\alpha=(a_1,\ldots,a_n)$, $\beta=(b_1,\ldots,b_n)$. But I am not sure what can I do from here. Any hint/comment is welcome!

RobPratt
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finiteness
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    Please see Sherman--Morrison formula: https://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula – m_gnacik Sep 07 '20 at 16:44

2 Answers2

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\begin{align*} (I + uv^\top)^{-1} &= I - uv^\top + (uv^\top)^2 - (uv^\top)^3 + \cdots \\ &= I - u(1 - v^\top u + (v^\top u)^2 - \cdots)v^\top \\ &= I - u \left(\frac{1}{1 + v^\top u} \right) v^\top \\ &= I - \frac{1}{1 + v^\top u} u v^\top. \end{align*}

This is a special case of the Sherman-Morrison Formula.

Trevor Gunn
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  • Your proof works only if $;\left|v^\top u\right|<1$, but the matrix $M$ is invertible for any $;v^\top u\ne -1.$ – Angelo Sep 07 '20 at 20:36
  • Who said anything about convergence? @Angelo I'm working with indeterminates. – Trevor Gunn Sep 07 '20 at 20:44
  • @Angelo use $2n$ inteterminates for the entries of $u$ and $v$. Take the ring $\mathbb{Z}[x_1,\dots,x_{2n}]$ localize at the ideal $I = (x_1,\dots,x_{2n})$ and then complete to get a power series ring $\mathbb{Z}[[x_1,\dots,x_{2n}]]$. Then because $v^\top u \in I$ and the ring is complete in the $I$-adic topology, it converges in $\mathbb{Z}[[x_1,\dots,x_{2n}]]$. Now you can tensor with whatever field you want and specialize $x_i = a_i$ and you get a true equation. But we don't really need to worry about completion and $I$-adic topology. The method "just works™." – Trevor Gunn Sep 07 '20 at 20:55
  • So, by using your method, we can tensor with whatever field we want and get that $;1-v^\top u + (v^\top u)^2 -\ldots =\cfrac{1}{1+v^\top u};$ holds for any value of the field, even for $v^\top u=10$, but I think that $1-10+100-\ldots=\cfrac{1}{11}$ is false. – Angelo Sep 07 '20 at 21:08
  • @Angelo You get an identity of rational functions whose proof involves power series. The power series converge formally. So the power series identity is true for any ring complete with respect to an ideal $I$ and for any element of $I$. But the identity of rational functions is true for any ring and any element such that the relevant denominators are non-zero. – Trevor Gunn Sep 07 '20 at 21:15
  • @Angelo See https://math.stackexchange.com/q/3599218/437127 and https://math.stackexchange.com/a/675128/437127 and https://mathoverflow.net/q/31595/111804 for discussion of this method. – Trevor Gunn Sep 07 '20 at 21:19
  • I saw your link, but the matrix $M$ contains numbers not indeterminates. – Angelo Sep 07 '20 at 21:27
  • According to what you say, the identity $;1-v^\top u+(v^\top u)^2-\ldots=\cfrac{1}{1+v^\top u};$ should be true for any element such that the denominator $;1+v^\top u\ne0;$, even for $v^\top u=10$. Actually $1-10+100-\ldots=\cfrac{1}{11}$ is false. – Angelo Sep 07 '20 at 21:29
  • @Angelo Make it contain indeterminate, get a formula that works and then substitute back. The power series identity doesn't make sense over the reals but it does make sense in the 2-adics or 5-adics for example. The rational function identity (which is what we need) is true in any ring even though the intermediate steps don't work in every ring. There is a formalism in which this works and I encourage you to open a new question about this if you are curious because it is a very interesting question. That's all I want to discuss about this here. – Trevor Gunn Sep 07 '20 at 21:32
  • The matrix $M$ does not contain $2$-adics or $5$-adics, it contains real or complex numbers. How can you say that the formula $;1-v^\top u+(v^\top u)^2-\ldots=\cfrac{1}{1+v^\top u};$ works even for $\left| v^\top u\right|\ge1$? Even if a formula holds with indeterminates in a ring, it does not mean that it also holds for any value of a field (apart those that vanish the denominators). – Angelo Sep 07 '20 at 21:47
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It is not necessary to use series or Sherman-Morrison Formula.

$M=I_n+\alpha^T\beta$

where $\alpha=(a_1,a_2,\ldots,a_n)$, $\beta=(b_1,b_2,\ldots,b_n)$ and $I_n$ is the identity matrix of order $n.\;$ Moreover, $\;1+\beta\alpha^T\ne0$.

We are looking for a matrix $N$ such that $MN=NM=I_n$.

$MN=I_n\iff\left(I_n+\alpha^T\beta\right)N=I_n\iff\\\iff N+\alpha^T\beta N=I_n\quad\color{blue}{(*)}$

and, by multiplying both sides of the last equality $(*)$ by the matrix $\;\alpha^T\beta\;,\;$ we get that

$\alpha^T\beta N+\alpha^T\left(\beta\alpha^T\right)\beta N=\alpha^T\beta\;,$

but $\;\beta\alpha^T$ is a number, hence it commutes with respect to multiplication,

$\alpha^T\beta N+\left(\beta\alpha^T\right)\alpha^T\beta N=\alpha^T\beta\;,$

$\left(1+\beta\alpha^T\right)\alpha^T\beta N=\alpha^T\beta\;.\quad\color{blue}{(**)}$

Since $\;1+\beta\alpha^T\ne0\;$, by dividing both sides of the last equality $(**)$ by the number $\;1+\beta\alpha^T$, we get that

$\alpha^T\beta N=\cfrac{1}{1+\beta\alpha^T}\alpha^T\beta $

and from $(*)$ it follows that

$N=I_n-\cfrac{1}{1+\beta\alpha^T}\alpha^T\beta\;.$

Moreover it is easy to verify that not only $MN=I_n$ but also $NM=I_n$, hence the matrix

$N=I_n-\cfrac{1}{1+\beta\alpha^T}\alpha^T\beta$

is the inverse of $\;M\;.$

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If $\;1+\beta\alpha^T=0\;$ then the matrix $M$ is not invertible.

Indeed, if $M$ were invertible, there would exist a matrix $N$ such that $MN=I_n$ and from $(**)$ it would follow that $\alpha^T\beta=0$ (zero matrix).

But $\;\alpha^T\beta=0\;$ implies that $\;\alpha\;$ or $\;\beta\;$ is the zero vector, hence $\;\beta\alpha^T=0\;$ and it contradicts the hypothesis $\;1+\beta\alpha^T=0.\;$ Consequently the matrix $M$ cannot be invertible otherwise it would lead to a contradiction.

Angelo
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