Convergence of an increasing, bounded sequence of positive definite matrices

Question

Given two symmetric $n \times n$ matrices $\mathsf{B}_1, \mathsf{B}_2$, we say $\mathsf{B}_1 \geq \mathsf{B}_2$ if $\mathsf{B}_1 - \mathsf{B}_2$ is positive semi-definite. We can easily see that this implies $x^\mathsf{T} \mathsf{B}_1 x \geq x^\mathsf{T} \mathsf{B}_2 x$ for all $x \in \mathbb{R}^n$.

Suppose we have a sequence of positive definite matrices $\{\mathsf{Q_\ell}\}_{\ell \geq 0}$ which is increasing and bounded with respect to this partial order. Does this sequence converge?

I'm reading a paper in which the authors use this claim, but I have not figured out how to justify it.

I can see that for each $x \in \mathbb{R}^n$, $\{x^\mathsf{T} \mathsf{Q_\ell} x\}_{\ell \geq 0}$ is an increasing sequence of real numbers which is bounded above, and is therefore convergent. Letting $x$ be the standard basis vectors of $\mathbb{R}^n$, we can see that the diagonal entries converge. We can then see that the off-diagonal entries are bounded by using the Cauchy-Schwarz inequality, but I cannot see why they must converge.

If you can argue that there is some positive definite matrix $\mathsf{Q}$ such that for any $x$: $ x^\mathsf{T} \mathsf{Q_\ell} x \rightarrow x^\mathsf{T} \mathsf{Q} x $, as $\ell \rightarrow \infty$, then we could conclude that $ \mathsf{Q_\ell} \rightarrow \mathsf{Q} $ by using the polarization identity:

$$ x^\mathsf{T} \mathsf{Q_\ell} y = \frac{1}{4} ( (x+y)^\mathsf{T} \mathsf{Q_\ell} (x+y) - (x-y)^\mathsf{T} \mathsf{Q_\ell} (x-y) ) $$

We know that the terms on the right-hand side converge, so the left-hand side must also converge for any $x, y \in \mathbb{R}^n$. Letting $x$ and $y$ be standard basis vectors, we could conclude that the individual matrix entries converge.

Answer 1

I managed to find the solution to this. I had posted all the necessary steps in the original question, but couldn't figure out the last piece of the puzzle for some reason.

For any $x \in \mathbb{R}^n$, the sequence of real numbers $\{ x^\mathsf{T} \mathsf{Q}_\ell x \}_{\ell \geq 0}$ is increasing and bounded above. Therefore the sequence $\{ x^\mathsf{T} \mathsf{Q}_\ell x \}_{\ell \geq 0}$ converges to its supremum for all $x \in \mathbb{R}^n$. To prove that the matrix entries converge, we can make use of the polarisation identity: $$ x^\mathsf{T} \mathsf{Q_\ell} y = \frac{1}{4} \left( (x+y)^\mathsf{T} \mathsf{Q_\ell} (x+y) - (x-y)^\mathsf{T} \mathsf{Q_\ell} (x-y) \right) \quad \text{for all } x, y \in \mathbb{R}^n $$

The terms on the right-hand side are of the form $ z^\mathsf{T} \mathsf{Q}_\ell z $, so they converge as $\ell \rightarrow \infty$. Therefore, the left-hand side must also converge for any $x,y$. If we let $x = \hat{e}_i$ and $y = \hat{e}_j$, we can see that the sequences of matrix entries $ (\mathsf{Q}_\ell)_{i,j} = \hat{e}_i^\mathsf{T} \mathsf{Q}_\ell \hat{e}_j $ must converge as $\ell \rightarrow \infty$ for all $i,j$.

Answer 2

Here is a sketch of an argument.

I will use $\succ$ to mean what you call $\ge$. One can check that for psd matrices, if both $A\succ B$ and $B\succ A$ then $A=B$.

Your boundedness condition implies that $B_n$ has convergent subsequences. Suppose $L_1$ and $L_2$ are two distinct limit points. By passing to subsequences you can assume $B_{2n}\to L_1$ and $B_{2n+1}\to L_2$. But $B_{2n+2}\succ B_{2n+1}\succ B_{2n}$, so $L_1\succ L_2$, and similarly $L_2\succ L_1$. But then $L_1=L_2$>