# If $A_n B_n$ converge to the identity matrix, is it true that $A_n^{1/2} B_n^{1/2} \to I$ as well?

If two sequences of positive definite symmetric matrices $$(A_n)$$ and $$(B_n)$$ are such that $$A_n B_n \to I$$, is it necessarily true that $$A_n^{1/2} B_n^{1/2}$$ also converge to the identity?

If not, what sort of additional assumptions are needed? (I can see that if $$A_n$$ and $$B_n$$ are simultaneously diagonalizable, the implication holds. Are there weaker conditions that also guarantee this?)

• Are we assuming that the matrices (and their square roots) are positive semidefinite or something? Otherwise, $A_n=B_n=I$ and $A_n^{1/2}=I, B_n^{1/2}=-I$ is an immediate counterexample. Apr 4, 2020 at 21:57
• Yes, thanks for that; for the application I have in mind, the sequences are non-singular covariance matrices. Apr 4, 2020 at 21:59
• If your matrices are uniformly bounded that seems to work. But I imagine that this is not the case in your situation Apr 4, 2020 at 22:13
• You mean norm convergence right? Apr 4, 2020 at 22:24

For any nonsingular matrix $$V$$ and Hermitian matrix $$H$$, if $$x$$ is an eigenvector of $$H$$ corresponding to the largest-sized eigenvalue, then $$\|H\|_2=\frac{\|V^{-1}HV(V^{-1}x)\|_2}{\|V^{-1}x\|_2}\le\|V^{-1}HV\|_2.$$ Let $$A_n^{1/2}B_n^{1/2}=U_nP_n$$ be a polar decomposition, where $$U_n$$ is unitary and $$P_n$$ is positive definite. Then $$P_n^2=(U_nP_n)^\ast(U_nP_n)\to I$$, because $$\|B_n^{1/2}A_nB_n^{1/2}-I\|_2 \le\|B_n^{-1/2}(B^{1/2}A_nB_n^{1/2}-I)B_n^{1/2}\|_2=\|A_nB_n-I\|_2\to0.$$ It follows that $$P_n\to I$$ too, because $$\|P_n-I\|_2= \|(P_n+I)^{-1}(P_n^2-I)\|_2\le\|(P_n+I)^{-1}\|\|P_n^2-I\|_2\le\|P_n^2-I\|_2\to0.$$ Now, if we can show that $$U_n\to I$$ then $$A_n^{1/2}B_n^{1/2}=U_nP_n\to I$$ and we are done.
For each $$n$$, let $$U_n=V_nZ_nV_n^\ast$$ be a unitary diagonalisation and let $$P_n-I=V_nH_nV_n^\ast$$. Since $$P_n\to I$$, we have $$H_n\to 0$$. Therefore, for any $$\epsilon>0$$, the absolute value of each entry of $$H_n$$ is smaller than $$\epsilon$$ when $$n$$ is sufficiently large. So, if $$Z_n$$ and $$H_n$$ are $$m\times m$$ matrices and if we denote by $$z_i$$ and $$h_{ij}$$ respectively the $$i$$-th diagonal entry of $$Z_n$$ and the $$(i,j)$$-th entry of $$H_n$$, then by Gerschgorin disc theorem, all eigenvalues of $$Z_n+Z_nH_n$$ lie inside the union of discs $$\bigcup_{i=1}^m D\left(z_i+z_ih_{ii},\sum_{j\ne i}|h_{ij}|\right),$$ where $$D(z,r)$$ denotes the closed disc on the Argand plane with center $$z$$ and radius $$r$$. However, it is well known that each connected component of some $$k$$ discs must contain exactly $$k$$ eigenvalues. Since the diameter of each disc is at most $$2(m-1)\epsilon$$ and there are $$m$$ discs in total, for each $$z_i$$ there exists some eigenvalue $$\lambda$$ of $$Z_n+Z_nH_n$$ such that $$|z_i-\lambda|\le 2m(m-1)\epsilon$$.
Yet, $$Z_n+Z_nH_n$$ is similar to $$U_nP_n=A_n^{1/2}B_n^{1/2}$$, which in turn is similar to the positive definite matrix $$A_n^{1/4}B_n^{1/2}A_n^{1/4}$$. Therefore $$\lambda>0$$ and \begin{aligned} z_i&\in\{|z|=1\}\cap\bigcup_{\lambda>0}\left\{|z-\lambda|\le2m(m-1)\epsilon\right\}\\ &=\{|z|=1\}\cap\{z=x+iy: |z|\le2m(m-1)\epsilon \text{ or } |y|\le2m(m-1)\epsilon\}. \end{aligned} As the intersection on last line shrinks to $$\{1\}$$ when $$\epsilon\to0$$, we see that the eigenvalues of $$Z_n$$ (i.e. the eigenvalues of $$U_n$$) approach $$1$$ when $$n\to\infty$$. Hence $$\|U_n-I\|_F^2=\sum_{i=1}^m|z_i-1|^2\to0$$, i.e. $$U_n\to I$$.
It should be clear in the special case that $$A_n\to A$$ for some psd matrix $$A$$.
And in the case that $$A_n/\|A_n\|\to A$$, since $$A_nB_n = (A_n/\|A_n\|)(B_n\|A_n|)$$.
And since the set of psd matrices $$M$$ for which $$\|M\|=1$$ is compact, a subsequence argument should finish the job.