Multivariate Application of Slutsky's lemma Let $(A_n)_{n \geq 1}$ be a sequence of random matrices in $\mathbb{R}^{d \times d}$ and $(T_n)_{n \geq 1}$ and $(Z_n)_{n \geq 1}$ be  two sequences of random vectors in $\mathbb{R}^{d}$. We assume that:


*

*$\forall n \geq 1$, $Z_n = A_n T_n$

*$A_n$ converges in probability towards a deterministic non singular random matrice $A$

*$Z_n$ converges in distribution towards a random vector $Z$.


Show that $T_n$ converges in distribution towards $A^{-1}Z$. We do not assume that the $A_n$ are non-singular.
If the $A_n$ were non-singular it would be easy to say $T_n = A_n^{-1}A_nT_n$ and apply Slutsky's lemma to obtain the result. Without assuming non-singularity but if we could assume $A_n$ nonnegative I would use $T_n = (A_n+ \frac{1}{n})^{-1}(A_n+ \frac{1}{n})T_n$ to show the result. In the general case, I do not know how to prove it. Any hint ? 
 A: Let $\|\cdot\|$ be some submultiplicative matrix norm. Since $GL_d(\mathbb R)$ is open, there is some $r>0$ such that $\overline B_{\|\cdot\|}(A,r)\subset GL_d(\mathbb R)$. It suffices to show that 
$$(1_{\|A_n-A\|\leq r}A_n^{-1},Z_n)\xrightarrow{d} (A^{-1},Z) $$ 
and $$T_n 1_{\|A_n-A\|> r} \xrightarrow{P} 0 $$
The continuous mapping theorem combined with Slutsky's theorem will then yield $$T_n = 1_{\|A_n-A\|\leq r}A_n^{-1} Z_n + T_n 1_{\|A_n-A\|> r}\xrightarrow{d} A^{-1}Z$$

Let $f:\mathbb R^{d\times d}\times \mathbb R^d \to \mathbb R$ be a $K$-Lipschitz function bounded by some $M$.
$$\begin{align}|E[f(1_{\|A_n-A\|\leq r}A_n^{-1},Z_n)-f(A^{-1},Z)]| \leq &|E[f(1_{\|A_n-A\|\leq r}A_n^{-1},Z_n)-f(A^{-1},Z_n)]|\\ 
+ &|E[f(A^{-1},Z_n) - f(A^{-1},Z)]|\end{align}$$
Since $A^{-1}$ is a constant, the mapping $z\mapsto f(A^{-1},z)$ is bounded and continuous, hence $|E[f(A^{-1},Z_n) - f(A^{-1},Z)]|\xrightarrow[n\to \infty]{} 0$.
Regarding the first summand, let $\epsilon >0$ and $E_n$ be the event $$\begin{align}E_n
:=& \left(\|(1_{\|A_n-A\|\leq r}A_n^{-1},Z_n)-(A^{-1},Z_n)\| \geq \epsilon \right)\\ 
=& \left(\|1_{\|A_n-A\|\leq r}A_n^{-1} - A^{-1}\|\geq \epsilon \right) 
\end{align}$$
$$\begin{align}|E[f(1_{\|A_n-A\|\leq r}A_n^{-1},Z_n)-f(A^{-1},Z_n)]|
&\leq K\epsilon  P(E_n) + 2MP(E_n^c)\\
&\leq K\epsilon + 2M P\left(\|1_{\|A_n-A\|\leq r}A_n^{-1} - A^{-1}\|\geq \epsilon \right)\tag{1}
\end{align}
$$
Note that $$\begin{align}P\left(\|1_{\|A_n-A\|\leq r}A_n^{-1} - A^{-1}\|\geq \epsilon \right)&= P\left(\|1_{\|A_n-A\|\leq r}(A_n^{-1}-A^{-1}) - A^{-1}1_{\|A_n-A\|>r}\|\geq \epsilon \right)\\
&\leq P\left(\|1_{\|A_n-A\|\leq r}(A_n^{-1}-A^{-1})\| \geq \frac \epsilon 2\right)+P\left(\|A^{-1}1_{\|A_n-A\|>r}\| \geq \frac \epsilon 2\right)\\
&\leq P\left(\|1_{\|A_n-A\|\leq r}(A_n^{-1}-A^{-1})\| \geq \frac \epsilon 2\right) + P(\|A_n-A\|>r )
\end{align}$$
The inverse function $\varphi:X\mapsto X^{-1}$ has Frechet derivative $d\varphi(X)(H)=-X^{-1}HX^{-1}$.
Suppose $A_n\in \overline B_{\|\cdot\|}(A,r)$ and consider some $B$ on the segment joining $A$ and $A_n$. Note that $\|df(B)\|_{op}\leq \|B^{-1}\|^2$ and the mapping $B\mapsto \|B^{-1}\|^2$ is continuous on the compact $\overline B_{\|\cdot\|}(A,r)$ so that there exists a constant $M'$ depending only on $A$ such that $\forall B\in [A_n,A], \|df(B)\|_{op}\leq M'$. The mean value inequality yields the estimate 
$$P\left(\|1_{\|A_n-A\|\leq r}(A_n^{-1}-A^{-1})\| \geq \frac \epsilon 2\right)\leq P\left(\|A_n-A\| \geq \frac {\epsilon}{ 2M'}\right)$$
It follows that $$P\left(\|1_{\|A_n-A\|\leq r}A_n^{-1} - A^{-1}\|\geq \epsilon \right)\xrightarrow[n\to \infty]{} 0$$ and taking $n\to \infty$ in $(1)$ we get
$\limsup_n |E[f(1_{\|A_n-A\|\leq r}A_n^{-1},Z_n)-f(A^{-1},Z_n)]|\leq K\epsilon$ for any $\epsilon>0$, hence 
$$|E[f(1_{\|A_n-A\|\leq r}A_n^{-1},Z_n)-f(A^{-1},Z_n)]| \xrightarrow[n\to \infty]{} 0$$
Finally,
$$(1_{\|A_n-A\|\leq r}A_n^{-1},Z_n)\xrightarrow{d} (A^{-1},Z) $$ 

For the last part, consider $\epsilon>0$ and note that 
$$P(T_n 1_{\|A_n-A\|> r} \geq \epsilon)  = P((\|T_n\|\geq \epsilon) \cap (\|A_n-A\|> r)) \leq P(\|A_n-A\|> r)$$

PS: V-E has a shorter solution, but quite out of the blue. Tell him that I say hi.
