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Let $\Sigma$ be a $k\times k$ positive semi-definite real matrix. Suppose the exists a sequence $(a_n)\subset\mathbb R^k$ with $|a_n|\leq M$ for all $n$ and $a\in\mathbb R^k$ such that

$$(a_n-a)^\top \Sigma (a_n-a) \to 0 \quad \text{as } n\to \infty$$

Question: Does there exists $b\in\mathbb R^k$ with $|b|\leq M$ such that $(a_n-b)^\top \Sigma (a_n-b) \to 0$ as $n\to \infty$?

If $\Sigma$ is positive definite with eigendecomposition $Q\Lambda Q^T$ then we get $(a_n-a)^\top Q\to 0$ which implies $a_n\to a$ and so we can take $b=a$.

But what about the case where some eigenvalues of $\Sigma$ are zero?

Alphie
  • 5,087
  • The answer is still yes. Note that $\Sigma$ is positive definite on $U=(\ker(\Sigma))^\perp$ and zero on $\ker(\Sigma)$. Therefore $x^T\Sigma x=u^T\Sigma u$ where $u$ is the component of $x$ on $U$. The given condition thus shows that the $u$-components of the $a_n-a$ tends to zero. Hence the $u$-component of $a$ has modulus bounded by $M$ and you may again take $b$ as the $u$-component of $a$. – user1551 Nov 15 '21 at 14:46
  • @user1551 Am not sure I see why the $u$ component of $a_n-a$ converges to zero. Can you explain? – Alphie Nov 15 '21 at 15:22
  • It's the same reason as in the positive definite case that you have explained in your question: call the $u$-component of $a_n-a$ as $u_n$. As $u_n^T\Sigma u_n=(a_n-a)^T\Sigma(a_n-a)\to0$, we get $\Sigma u_n\to0$ and (since $\Sigma$ is nonsingular on $U$,) $u_n\to0$. – user1551 Nov 15 '21 at 15:42
  • @user1551 If I understand correctly we have $\mathbb R^n =U\oplus U^\perp$ and $\Sigma$ splits into two linear maps $\Sigma_U: U\to U$ and $\Sigma_{U^\perp}: U^\perp\to U^\perp$ when restricted to $U, U^\perp$. Since $\Sigma_U$ is positive definite it has a positive square root $\Sigma^{1/2}_U$. Then from $u_n^\top\Sigma_U u_n= |\Sigma_U^{1/2}u_n|^2\to 0$ we get $u_n=\Sigma_U^{-1/2}\Sigma_U^{1/2}u_n \to0$ by continuity of linear maps on finite dimensional spaces. Is this correct? – Alphie Nov 15 '21 at 16:15
  • Yes, that should work. – user1551 Nov 15 '21 at 16:22
  • @user1551 Thanks a lot for your help. In my original setting $\Sigma$ is a random matrix , and I would like to project in measurable way, i.e. having $b$ measurable. I asked the question here: https://math.stackexchange.com/q/4306971/522332. If you have any ideas this would be very appreciated. – Alphie Nov 15 '21 at 17:27

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