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?)

  • 2
    $\begingroup$ 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. $\endgroup$
    – Arthur
    Apr 4, 2020 at 21:57
  • $\begingroup$ Yes, thanks for that; for the application I have in mind, the sequences are non-singular covariance matrices. $\endgroup$
    – Kevin
    Apr 4, 2020 at 21:59
  • $\begingroup$ If your matrices are uniformly bounded that seems to work. But I imagine that this is not the case in your situation $\endgroup$
    – H. H. Rugh
    Apr 4, 2020 at 22:13
  • $\begingroup$ You mean norm convergence right? $\endgroup$ Apr 4, 2020 at 22:24

2 Answers 2


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.


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