Upper bound on the trace of the inverse of a rank-1 perturbed symmetric matrix I have a rank-1 perturbed matrix of the form:
$\boldsymbol{H}_{t+1} = \boldsymbol{H}_{t} + xx^\top$
where $x \in \mathcal{X} \subset \mathbb{R}^{d}$, $\boldsymbol{H}_{t}$ is a symmetrix positive definite matrix. The vector $x$ is chosen to minimize the trace of $\boldsymbol{H}_{t+1}^{-1}$, i.e,
$x = \underset{x \in \mathcal{X}}{\operatorname{argmin}}  Tr(\boldsymbol{H}^{-1}_{t+1})$. For the sake of this question, you can assume that $\mathcal{X}$ is a unit hypercube in d dimensions. I want to show that sequentially applying this, the $Tr(\boldsymbol{H}^{-1}_{t})$ will converge to 0 as $t \to \infty$ assuming $\boldsymbol{H_{0}} = I_{d}$.
My Approach
I tried applying the sherman morrison formula, which gives the following:
\begin{equation}
\begin{split}
\boldsymbol H^{-1}_{t+1}(x) = \boldsymbol H^{-1}_{t} - \dfrac{\boldsymbol H^{-1}_{t} xx^{T}\boldsymbol H^{-1}_{t}}{1+x^{T}\boldsymbol H^{-1}_{t}x}
\end{split}
\end{equation}
and since the trace is a linear operator I can just do:
\begin{equation}
\begin{split}
Tr(\boldsymbol H^{-1}_{t+1}(x)) = Tr(\boldsymbol H^{-1}_{t}) - Tr(\dfrac{\boldsymbol H^{-1}_{t} xx^{T}\boldsymbol H^{-1}_{t}}{1+x^{T}\boldsymbol H^{-1}_{t}x})
\end{split}
\end{equation}
Now  you can see that the Trace of $\boldsymbol H^{-1}_{t+1}$ decreases strictly monotonically since $Tr(\boldsymbol H^{-1}_{t} xx^{T}\boldsymbol H^{-1}_{t}) = \Vert \boldsymbol H^{-1}_{t} x \Vert_{2}^{2}$, and since $x$ is chosen to maximize $\dfrac{\Vert \boldsymbol H^{-1}_{t} x \Vert_{2}^{2}}{1+x^{T}\boldsymbol H^{-1}_{t}x}$,  the term $Tr(\dfrac{\boldsymbol H^{-1}_{t} xx^{T}\boldsymbol H^{-1}_{t}}{1+x^{T}\boldsymbol H^{-1}_{t}x})$ is guaranteed to be positive. However, I can't manage establish a recursive formula to show that I can analyse its convergence. Preferebly, I am looking for something of the form:
\begin{equation}
\begin{split}
Tr(\boldsymbol H^{-1}_{t+1}(x)) = Tr(\boldsymbol H^{-1}_{t})* \alpha(t)
\end{split}
\end{equation}
where $\alpha(t)$ is a factor smaller than 1 and can depend on $t$. Any thoughts?
 A: $
\def\a{\alpha}\def\b{\beta}\def\l{\lambda}
\def\H{H^{-1}}\def\HH{H^{-2}}
\def\o{{\tt1}}\def\p{\partial}
\def\LR#1{\left(#1\right)}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
\def\gradLR#1#2{\LR{\grad{#1}{#2}}}
$After applying the Sherman-Morrison formula you can write the problem as
$$\eqalign{
\max_x \fracLR{x^T\HH x}{\o+x^T\H x} \\
}$$
where the subscript on $H_t$ has been dropped for typing convenience.
Now you can follow the usual prescription of calculating the gradient, setting it to zero, and solving for the optimal $x$ vector.
Start by defining two scalar variables
$$\eqalign{
\a &= x^T\HH x &\qiq \gradLR{\a}{x} = 2\HH x \\
\b &= x^T\H x+\o &\qiq \gradLR{\b}{x} = 2\H x \\
}$$
and use these to calculate the gradient of the objective function
$$\eqalign{
\l &= \b^{-1}\a \\
\grad{\l}{x} &= \b^{-1}\gradLR{\a}{x} \;-\; \a\b^{-2}\gradLR{\b}{x} \\
0 &= 2\b^{-1}\LR{\HH x - \l\H x} \\
\l x &= \H x \\
}$$
Note that this is just the standard eigenvalue problem.
You want the maximum eigenvalue of $\H$ which is equal to
the minimum eigenvalue of $H$.
The optimal $x$ vector is the associated eigenvector.
