limit of a matrix product I'm trying to show that the limit of a particular matrix product approaches another matrix. The first is of the form:
$$G =mA\Bigl(B+(A'A)m\Bigr)^{-1}$$
The second is of the form:
$$H=(AB^{-1}A')^{-1}AB^{-1}$$
Where $m$ is a scalar, $A$ is a $k\times n$ matrix with $k << n$ and $A'A$ is singular. $B$ is a positive definite symmetric matrix.  Both $G$ and $H$ are well defined. I want to show the following:
$$\lim_{m\rightarrow \infty}G=H$$
Numerically this limit works but I'm not sure how to prove it. I've tried various tricks with factoring and I've thought that I may have to resort to very carefully writing out the matrix products and examining terms but would really like to avoid this.  Any help would be greatly appreciated
thanks
 A: Let us write
$$G_{m} = G = mA (B + mA^{*}A)^{-1} $$
in order to clarify that $G_{m}$ depends on $m$. Then we claim that $G_{m} \to H$ in operator norm. To this end, we first check the following two simple lemmas:

Lemma 1. Let $J = (AB^{-1}A^{*})^{-1}$. Then we have
  $$ G_{m}
= J \left( AB^{-1} - \frac{1}{m}G_{m} \right)
= H - \frac{1}{m} JG_{m}. \tag{1} $$

Proof. We consider $J^{-1}G_{m} = AB^{-1}A^{*}G_{m}$ instead. Simplifying this, we get
\begin{align*}
AB^{-1}A^{*}G_{m}
&= (AB^{-1}A^{*})(mA)(B + mA^{*}A)^{-1} \\
&= AB^{-1}(mA^{*}A)(B + mA^{*}A)^{-1} \\
&= AB^{-1}\{(B + mA^{*}A) - B\}(B + mA^{*}A)^{-1} \\
&= AB^{-1}\left( I - B(B + mA^{*}A)^{-1} \right) \\
&= AB^{-1} - AB^{-1}B(B + mA^{*}A)^{-1} \\
&= AB^{-1} - \frac{1}{m}G_{m}.
\end{align*}
Multiplying $J$ to both sides, we obtain $\text{(1)}$ as desired. ////

Lemma 2. For $m > \| J \|$, the matrix $1 + m^{-1}J$ is invertible and we have the following estimate
  $$ \left\| (I + m^{-1}J)^{-1} - I \right\| \leq \frac{\| J \|}{m - \| J \|}. \tag{2} $$

Proof. Let $u \in \Bbb{R}^{n}$ be any vector. Then
$$ \| (I + m^{-1}J) u \| \geq \| u \| - \| m^{-1} J u \| \geq (1 - m^{-1}\| J \|) \| u \|. $$
In particular, if $m^{-1}\| J \| < 1$, this shows that $I + m^{-1}J$ is injective, hence bijective. This proves that $I + m^{-1}J$ is invertible. Moreover, in this case we can replace $u$ by $(I + m^{-1}J)^{-1} u$ and we get
$$ \left\| (I + m^{-1}J)^{-1} u \right\| \leq \frac{1}{1 - m^{-1}\| J \|} \| u \|
\quad \Longrightarrow \quad
\left\| (I + m^{-1}J)^{-1} \right\| \leq \frac{1}{1 - m^{-1}\| J \|}. \tag{3} $$
On the other hand, we have
$$ \left\{ (I + m^{-1}J)^{-1} - I \right\} (I + m^{-1}J)
= I - (I + m^{-1}J)
= -m^{-1}J. \tag{4} $$
Thus combining $\text{(3)}$ and $\text{(4)}$, it follows that
$$ \left\| (I + m^{-1}J)^{-1} - I \right\|
= \left\| -m^{-1} J (I + m^{-1}J)^{-1} \right\|
\leq \frac{\| J \|}{m} \frac{1}{1 - m^{-1} \| J \|}. $$
Therefore we get the estimate $\text{(2)}$ as desired. ////

Remark. Lemma 2 is a direct consequence of the Neumann series for $(1 + m^{-1}J)^{-1}$.


Now we return to the proof. 

Proposition. $G_{m} \to H$ in operator norm as $m \to \infty$.

Proof. Rearranging $\text{(1)}$, we get
$$ (I + m^{-1}J) G_{m} = H. $$
This if $m$ is large, then $G_{m} = (I + m^{-1}J)^{-1} H$ and hence we get
$$  
\| G_{m} - H \|
\leq \left\| (I + m^{-1}J)^{-1} - I \right\| \| H \|
\leq \frac{\| J \| \| H \|}{m - \|J\|}
\xrightarrow[]{m\to\infty} 0. $$
Therefore the proposition follows. ////
