1
$\begingroup$

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?

$\endgroup$

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.