Convergence of the positive operator $ \big( A+ (\lambda)^n B\big)^{\frac{1}{n}},$ where $0< \lambda <1.$

Question

Let $A, B$ be two positive operators in $M_{n}(\mathbb{C}).$ We know that $\lim_{n\to \infty}A^{\frac{1}{n}}$ converges and $\lim_{n\to \infty}A^{\frac{1}{n}}= P_{\rm ran(A)},$ where $P_{\rm ran(A)}$ denote the projection into range of $A.$ I am interested to know the limit of the sequence $$\lim_{n\to \infty} \big( A+ (\lambda)^n B\big)^{\frac{1}{n}},$$ where $0< \lambda <1.$ Some simple case, we can conclude that the above limit exists. However, I do not know in full generality. Any suggestions and input will be highly appreciated.

Answer 1

The limit exists and is equal to $\Pi_{\operatorname{range}(A)}+\lambda\Pi_{\operatorname{range}(B)\cap\ker(A)}$, where $\Pi_S$ denotes the orthogonal projection onto a linear subspace $S$.

The result is clear if $A=0$ or $B=0$. Let us assume that $A,B\ne0$. Since convergence of the sequence and the limit value are invariant under scaling of $A$ and $B$ by a common positive factor, we may further assume that $\|A+B\|_2\le1$.

Let $u$ be an arbitrary unit vector in $\operatorname{range}(A)$. Then $$ u^\ast A^{1/n}u \le u^\ast(A+\lambda^nB)^{1/n}u \le u^\ast(A+B)^{1/n}u \le \|(A+B)^{1/n}\|_2 \le 1. $$ Pass $n$ to the limit, we obtain $$ 1=u^\ast u=u^\ast\Pi_{\operatorname{range}(A)}u \le \lim_{n\to\infty}u^\ast(A+\lambda^nB)^{1/n}u \le 1. $$ Hence $$ \lim_{n\to\infty}u^\ast(A+\lambda^nB)^{1/n}u=1\tag{1} $$ or equivalently, $\lim_{n\to\infty}\langle u,u_n\rangle=1$ where $u_n=(A+\lambda^nB)^{1/n}u$. Since $\|u_n\|_2\le\|(A+\lambda^nB)^{1/n}\|_2\le\|(A+B)^{1/n}\|_2\le1$, we must have $$ \lim_{n\to\infty}u_n=u.\tag{1a} $$

Now let $v$ be an arbitrary unit vector in $\operatorname{range}(B)\cap\ker(A)$. Then $$ \lim_{n\to\infty}v^\ast(A+\lambda^nB)^{1/n}u=v^\ast \lim_{n\to\infty}v^\ast u_n=v^\ast u=0.\tag{2} $$ Furthermore, for any fixed $n$, if $A+\lambda^n B=USU^\ast$ is a unitary diagonalisation and we put $p=U^\ast v$, then $$ \begin{aligned} \lambda v^\ast B^{1/n}v &\le v^\ast(A+\lambda^n B)^{1/n}v\\ &=p^\ast S^{1/n}p\\ &=\sum_i|p_i|^2s_i^{1/n} \quad\text{(which is a convex combination of the $s_i^{1/n}$s)}\\ &\le(\sum_i|p_i|^2s_i)^{1/n} \quad\text{(because $x\to x^{1/n}$ is concave on $\mathbb R_{\ge0}$)}\\ &=(p^\ast Sp)^{1/n}\\ &=\left(v^\ast(A+\lambda^n B)v\right)^{1/n}\\ &=\lambda(v^\ast Bv)^{1/n} \quad\text{(because $Av=0$)}\\ &\le\lambda\|B\|_2^{1/n}. \end{aligned} $$ Pass $n$ to the limit, we obtain $\lambda=\lambda v^\ast \Pi_{\operatorname{range}(B)}v \le \lim_{n\to\infty}v^\ast(A+\lambda^n B)^{1/n}v \le \lambda$. Therefore $$ \lim_{n\to\infty}v^\ast(A+\lambda^n B)^{1/n}v=\lambda.\tag{3} $$ Finally, we also have $\ker(A)\cap\ker(B)\subseteq\ker(A+\lambda^n B)=\ker\left((A+\lambda^n B)^{1/n}\right)$. Therefore $$ \lim_{n\to\infty}x^\ast(A+\lambda^n B)^{1/n}w=0\tag{4} $$ for any $w\in\ker(A)\cap\ker(B)$ and any vector $x$. It follows from $(1)-(4)$ and the polarisation identity that $\lim_{n\to\infty}x^\ast(A+\lambda^n B)^{1/n}y$ exists for every pair of vectors $x$ and $y$. Consequently, $L=\lim_{n\to\infty}(A+\lambda^n B)^{1/n}$ exists.

Now let $U=\operatorname{range}(A),\,V=\operatorname{range}(B)\cap\ker(A)$ and $W=\ker(A)\cap\ker(B)$. The whole vector space is then the orthogonal sum of $U,V$ and $W$. By $(1a)$, we have $Lu=u$ for every $u\in U$. Therefore $U$ is an invariant subspace of $L$ and $L|_U=\operatorname{id}$.

By $(2)$ and $(4)$ (with $x\in V$), we also have $LV\perp U$ and $LV\perp W$. Hence $LV\subseteq(U+W)^\perp=V$, i.e., $V$ is an invariant subspace of $L$. By $(3)$, $v^\ast Lv=\lambda$ for every $v\in V$. Hence $L|_V=\lambda\operatorname{id}$.

Now $(4)$ also implies that $LW=0$. Hence $W$ is an invariant subspace of $L$ and $L|_W=0$.

It follows that $L=\Pi_U+\lambda\Pi_V=\Pi_{\operatorname{range}(A)}+\lambda\Pi_{\operatorname{range}(B)\cap\ker(A)}$.