I'm not a mathematician, so please have mercy on me.
Let $P_n$ denote the set of $n\times n$ postive semidefinite symmetric matrices with values in $\mathbb R$. Assume I have a sequence of matrices $(R_i)_{i=1}^{\infty}$ with elements in $P_n$. If for some $N\in\mathbb N$ I consider the partial sum $$ \tilde S_N := \sum_{i=1}^N R_i,\qquad (1) $$ then, since $R_i\in P_n$ so is $\tilde S_n$. Moreover I can state (?) that the sequence $\{\tilde S_j\}_{j=1}^N$ is rank-increasing (namely $rank \tilde S_j\ge rank \tilde S_k$ for all $k\le j$). However (1) might not converge in $P_n$ in the usual 2-matrix norm.

Thus, instead of (1), for any $N\in\mathbb N$ I consider the weighted sum $$ S_N := \sum_{i=1}^N \mu^{N-i} R_i \qquad (2) $$ with $\mu\in(0,1)$. Noting that we have $S_k = \mu S_{k-1} + R_k$ for all $k\in\{1,\dots,N\}$ and with $S_0:=0$, I have that for all $k=1,\dots,N$, $S_N\in P_n$ and the sequence $\{S_i\}_{i=1}^N$ is rank increasing (?).

If the matrices $R_i$ are uniformly bounded, in the sense that there exists $M>0$ such that $\|R_i\|_2\le M$ for all $i\in\mathbb N$, and since $P_n$ is closed, then I can conclude that (2) converges to a matrix $S_\infty\in P_n$ as $N$ grows to infinity.

Now let define for each $i$ the matrix $B_N:=(S_N)^\dagger$, with $\cdot^\dagger$ denoting the Moore-Penrose pseudoinverse. Is it true that $$ B_N \to_{N\to\infty} (S_\infty)^\dagger $$ ?
On one side I would say yes, since the sequence $\{S_N\}_N$ is rank increasing, thus there exists $m$ such that for all $i,j\ge m$ we have $rank S_i = rank S_{j}$.

However, on the other side there is something not working. Consider for instance $n=2$ and a sequence of matrices $R_i$ defined as $$ R_1= \begin{pmatrix}1&0\\0&1\end{pmatrix},\quad R_i=\begin{pmatrix}1&0\\0&0\end{pmatrix},\; i>2 $$ Then $R_1$ in the sum $S_N$ is weighted with $\mu^{N-1}$. Thus as $N$ grows its weight tends to zero. Therefore it seems that also if all $S_N$ have rank 2, $S_\infty$ has rank 1...

What do you think?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.