I have the following recurrence relation: $$ \alpha \times A_{t+1} = \beta \times A_{t} + uu^{T}$$ where $A = n \times n, u = n \times 1 $ and $\alpha , \beta$ are scalars. I would like to apply the Woodbury Identity to this relation if possible:
$$ (A + UCV)^{-1} = A^{-1} - A^{-1}U(C^{-1} + VA^{-1}U)^{-1} VA^{-1}$$ My first guess was to rearrange the rhs to make it look like the lhs of the Woodbury identity, something like:
$$ \frac{\alpha}{\beta} A_{t+1} = A_{t} + \frac{uu^{T}}{\beta} $$
My idea was to make $\beta$ the $1 \times 1$ C-Matrix from the Woodbury identity, so far this approach is incorrect. In the end I would like to find $ (A_{t+1})^{-1} $.
Every hint or suggestion is appreciated, thanks!
I assume that $\alpha$ and $\beta$ are non-zero. Note that $$ A_{t+1}^{-1} = \left(\frac1{\alpha}\cdot (\beta A_t + uu^T)\right)^{-1} = \alpha (\beta A_t + uu^T)^{-1}. $$ With the Sherman-Morrison formula (which is the same as applying the Woodbury formula in the case that $U^T,V$ are column vectors), we have $$ \begin{align} (\beta A_t + uu^T)^{-1} &= (\beta A_t + uu^T)^{-1} \\ & = (\beta A_t)^{-1} + \frac{(\beta A_t)^{-1}uu^T (\beta A_t)^{-1}}{1 + u^T (\beta A_t)^{-1}u} \\ & = \beta^{-1}A_t^{-1} + \frac{\beta^{-2} A_t^{-1}uu^T A_t^{-1}}{1 + \beta^{-1} u^T A_t^{-1}u} \\ & = \beta^{-1}A_t^{-1} + \beta^{-1}\frac{A_t^{-1}uu^T A_t^{-1}}{\beta + u^T A_t^{-1}u} = \beta^{-1}\left[ A_t^{-1} + \frac{A_t^{-1}uu^T A_t^{-1}}{\beta + u^T A_t^{-1}u} \right]. \end{align} $$