PD matrix perturbation I ran into the following problem in my research:
Given an invertible matrix $A \in \mathbb{C}^{n\times n}$, and define $B = A^{-1}$. Denote the $i$th standard basis in $\mathbb{R}^n$ by $e_i$ and the $i$th row of $B$ by $B_{i\bullet}$. Then is the following true
\begin{equation} \label{1}
\alpha A^HA - e_ie_i^T \succeq 0,
\end{equation}
when $\alpha \geq \|B_{i\bullet}\|_2^2$?
Numerical results seem to suggest that it always holds but I've not been able to find a proof.
 A: What I thought is the following: let $z$ be a complex nonzero  columnvector, then we want to show that $$z^H(\alpha A^HA - e_ie_i^T)z \geq 0.$$
This is equal with showing that 
$$\alpha (Az)^H(Az) - z^He_ie_i^Tz \geq 0.$$
We have that the $2$-norm $\|x\|_2^2 = \langle x, x \rangle$, where $\langle \cdot, \cdot \rangle$ denotes the inner product of vectors. Moreover, you can compute yourself that $z^He_ie_i^Tz = |z_i|^2$, where $z_i$ is the $i$th entry of the vector $z$.
Let us now focus on the term $\alpha(Az)^H(Az)$ for which we have the following inequality:
$$\alpha(Az)^H(Az) \geq \|B_{i\bullet}\|_2^2 \|Az\|_2^2$$
because of the definition of $\alpha$. 
Let us now use the inequality of Cauchy-Schwarz, which states that for every $p$-norm we have that $|x^Hy| \leq \|x\|_p\|y\|_p$. Applying this for $p = 2$ and the fact that $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^2$ is a strictly increasing function on $\mathbb{R}^+$, we have that 
$$\|B_{i\bullet}^H\|_2^2 \|Az\|_2^2 \geq |B_{i\bullet} Az|^2$$
(we can write $\|B_{i\bullet}^H\|_2^2 = \|B_{i\bullet}\|_2^2$ for a vector).
If we now note that $Az$ is a linear combination of the columns of $A$ with coefficients $z_j$ (the entries of the vector $z$) and we also have that $B_{i\bullet}a_j = \delta_{i,j}$ where $a_i$ denotes the $i$th column of the matrix $A$, we have that 
$$|B_{i\bullet} Az|^2 = |B_{i\bullet}a_iz_i|^2 = |z_i|^2.$$
As a result, we find that 
$$\alpha (Az)^H(Az) - z^He_ie_i^Tz \geq 0 \geq |z_i|^2 - |z_i|^2 = 0,$$
which concludes this proof.

$\textbf{Remark:}$ Any thoughts on this proof are welcome, since I am not completely sure about the step where I say that the linear combination only has a term in $z_i$ since the other terms would vanish.
