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Here is one of the most famous equation called Sherman–Morrison formula (1951) when we want to get an inverse matrix.

$$(A+vw^{\text{T}})^{-1}=A^{-1}-\cfrac{A^{-1}vw^{\text{T}}A^{-1}}{1+{w}^{\text{T}}A^{-1}v}$$

Then, we consider to get the inverse matrix $(A+\lambda E)^{-1}$ where $E$ is an identical matrix and $\lambda$ is scalar. This is my opinion but I suppose that identical matrix can be expressed using vector $\bf{e}$ which is defined below.

$$\bf{e}=[\it{e_{\rm{1}}} \it{e_{\rm{2}}} \dots \it{e_{\rm{i}}} \dots \it{e_{\rm{n}}}\rm{]^{\text{T}}}$$

The vector has a characteristic which can be described to multiply another identical vector.

$$ E=\bf{e e^{\text{T}}}=[\it{e_{i}e_{j}}\rm{]} \ \ Then \ \ \ \rm{ \begin{cases} 1 \ \ (\it{i=j}\rm{)}\\ \\ 0 \ \ (i \not= j) \end{cases} } $$

Therefore, $(A-\lambda E)^{-1}$ can be evaluated below using the first formula and my definition.

$$(A+\lambda \bf{e e^{\text{T}}}\rm{)}^{-1}=\it{A}^{\rm{-1}}-\cfrac{\it{A}^{\rm{-1}}\lambda\bf{e e^{\text{T}}}\it{A^{\rm{-1}}}}{\rm{1}+\lambda\bf{e^{\text{T}}}\it{A}^{\rm{-1}}\bf{e}}$$

Thus, $$\it{A}^{\rm{-1}}\bf{e e^{\text{T}}}\it{A^{\rm{-1}}}=A^{\rm{-2}}$$

$$\displaystyle \bf{e^{\text{T}}}\it{A}^{\rm{-1}}\bf{e}=\sum_{\it{i,j}=\rm{0}}^{\rm{n}}\it{\widetilde{a}_{i,j}e_{i}e_{j}}=\sum_{\it{i}=\rm{0}}^{\rm{n}}\it{\widetilde{a}_{i,i}}=\rm{Tr}(\it{A^{\rm{-1}}}\rm{)}$$

Eventually,

$$(A+\lambda E)^{-1}=A^{-1}-\cfrac{\lambda A^{\rm{-2}}}{1+\lambda\rm{Tr}(\it{A}^{\rm{-1}}\rm{)}}$$

But, this suggestion is totally wrong. In addition the similar question already submitted here. Although, I don't know why is wrong about my consideration.

Thanks.

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Hayabusananji
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1 Answers1

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It's not at all clear what you mean by $$ \mathbf{e}=\begin{bmatrix} e_1 & e_2 & \dots & e_i & \dots & e_n \end{bmatrix} $$ If by $e_i$ you mean the $i$-th column of the identity, then you have just rewritten $E$ in a different way and, of course, $\mathbf{e}\mathbf{e}^T=E$, but also $$ \mathbf{e}^TA^{-1}\mathbf{e}=A^{-1} $$ and is not a scalar, so you can't use it in the denominator of the formula, which assumes the matrix to find the inverse of has the form $$ A+vw^T $$ where $v$ and $w$ are column vectors so, in particular, $vw^T$ is a rank $1$ matrix (the case $vw^T=0$ is trivial).

For $n>1$ there are no column vectors $v$ and $w$ such that $vw^T=E$, because $E$ has rank $n>1$: you may indeed use the formula in the case $n=1$, which is however so simple that it's not necessary to do any calculation.

egreg
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