# Find the inverse of the following $n \times n$ matrix

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!

\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.

• Your proof works only if $\;\left|v^\top u\right|<1$, but the matrix $M$ is invertible for any $\;v^\top u\ne -1.$ Sep 7, 2020 at 20:36
• Who said anything about convergence? @Angelo I'm working with indeterminates. Sep 7, 2020 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™." Sep 7, 2020 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. Sep 7, 2020 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. Sep 7, 2020 at 21:15

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.