Bounds on a function of a positive definite matrix Suppose that $A$ is an $n\times n$ positive definite matrix and let $A^{(i)}$ denote the block corresponding to the first $i$ rows and columns of $A$. In addition, let $\sigma_1^2=A_{11}$ and
$$
\sigma_i^2=A_{22}^{(i)}-A^{(i)}_{21}\big[A_{11}^{(i)}\big]^{-1}A^{(i)}_{12}, \quad i>1,
$$
where $A_{kl}$ is the $kl$-th block of $A^{(i)}$ partitioned as follows:
$$
 A^{(i)}\equiv \begin{bmatrix}
  A_{11}^{(i)} & A_{12}^{(i)} \\
  A_{21}^{(i)} & A_{22}^{(i)}
 \end{bmatrix} \quad\text{with sizes}\quad
 \begin{bmatrix}
  (i-1)\times (i-1) & (i-1)\times 1 \\
  1\times (i-1) & 1\times 1
 \end{bmatrix}.
$$
I am looking for lower and upper bounds on
$$
\bbox[cornsilk,5px]
{
\psi(A)=\sum_{i=1}^n\frac{1}{\sigma_i}
}
$$
in terms of the eigenvalues of $A$.

It is obvious that $\psi(A)\ge \sigma_1^{-1}$. If $Q\Lambda Q^{\top}$ is the eigendecomposition of $A$ with $Q^{\top}Q=I$, then
$$
\sigma_1^2=e_1^{\top}A e_1=\sum_{i=1}^n [Q_{i,1}]^2\Lambda_{i,i}\le \lambda_{\text{max}},
$$
where $Q_{i,j}$ is the $(i,j)$-th element of $Q$ and $\lambda_{\max}=\max_{1\le i\le n}\Lambda_{i,i}$. Thus,
$$
\psi(A)\ge \frac{1}{\sqrt{\lambda_{\text{max}}}}.
$$
Is it possible to establish a similar upper bound on $\psi(A)$?
 A: I make this an answer because it's too much for a comment. There may be better estimates. By $\lambda_{\min}$/$\lambda_{\max}$ I denote the smallest/largest eigenvalue of $A$ and by $\lambda_{\min}^{(i)}$/$\lambda_{\max}^{(i)}$ the smallest/largest eigenvalue of $A^{(i)}$. Note that for a vector $x\in\mathbb R^i$ we have
$$
x^TA^{(i)}x = [x^T\;0]A\left[\begin{matrix}x\\0\end{matrix}\right]\ge\lambda_{\min}\|x\|^2.
$$
Therefore, $\lambda_{\min}^{(i)} = \min_{\|x\|=1}x^TA^{(i)}x\ge\lambda_{\min}$. Similarly, $\lambda_{\max}^{(i)} \le\lambda_{\max}$.
We have $\sigma_1^2 = e_1^TAe_1\ge\lambda_{\min}$. Now, let $i\ge 2$ and put
$$
N := \left[\begin{matrix}I & -[A_{11}^{(i)}]^{-1}A_{12}^{(i)}\\0 & 1\end{matrix}\right]
\qquad\text{and}\qquad
D := \left[\begin{matrix}A_{11}^{(i)} & 0\\0 & \sigma_i^2\end{matrix}\right].
$$
Then $D = N^TA^{(i)}N$ (compute it!). Hence, with $v := N[0\;1]^T$,
$$
\sigma_i^2 = [0\;1]D\left[\begin{matrix}0\\1\end{matrix}\right] = v^TA^{(i)}v\,\ge\,\lambda_{\min}^{(i)}\|v\|^2\,\ge\,\lambda_{\min}\cdot (1 + \|[A_{11}^{(i)}]^{-1}A_{12}^{(i)}\|^2)\,\ge\,\lambda_{\min}.
$$
Hence, $\psi(A)\le\tfrac n{\sqrt{\lambda_{\min}}}$. But also
$$
\sigma_i^2 = v^TA^{(i)}v\le\lambda_{\max}^{(i)}\|v\|^2\le\lambda_{\max}(1 + \|[A_{11}^{(i)}]^{-1}A_{12}^{(i)}\|^2)\le\lambda_{\max}(1+\|[A_{11}^{(i)}]^{-1}\|^2\|A_{12}^{(i)}\|^2).
$$
Now,
$$
\|[A_{11}^{(i)}]^{-1}\| = \|[A^{(i-1)}]^{-1}\| = [\lambda_{\min}^{(i-1)}]^{-1}\le\lambda_{\min}^{-1}
$$
and $\|A_{12}^{(i)}\|^2\le\|Ae_i\|^2\le\lambda_{\max}^2$. Set $\nu := \tfrac{\lambda_{\max}}{\lambda_{\min}}$ (the condition number of $A$). Then, setting everything together, we get
$$
\psi(A) = \frac 1{\sigma_1} + \sum_{i=2}^n\frac 1{\sigma_i}\,\ge\,\frac 1{\sqrt{\lambda_{\max}}} + \frac{n-1}{\sqrt{\lambda_{\max}}\sqrt{1+\nu^2}}.
$$
