Let $\Sigma_n$ be a sequence of $n\times n$ (growing size) positive definite matrices and suppose that $\lambda^{1}_{n} \to 0$, where $\lambda^{1}_{n}$ denotes the largest eigenvalue of $\Sigma_n$.

Using the eigendecomposition of $\Sigma_n$ we have that $\left\lVert \Sigma_n \right\rVert_F \leq \sqrt{n} \lambda^{1}_{n}$ so I cannot deduce that $\left\lVert \Sigma_n \right\rVert_F \to 0$, where $\left\lVert \cdot \right\rVert_F$ denotes the Frobenius norm.

But if $Q_n$ is an arbitrary bounded sequence of $n\times n$ (growing size) matrices then I can obtain the following inequality:

$$ \left\lVert \Sigma_n Q_n \right\rVert_F \leq \left\lVert \Sigma_n \right\rVert_2 \left\lVert Q_n \right\rVert_F = \lambda^{1}_{n} \left\lVert Q_n \right\rVert_F \to 0 $$ where $\left\lVert \cdot \right\rVert_2$ denotes the 2-norm.

What is going on here? Why is the matrix sequence $\Sigma_n$ not converging to zero but converges to zero when multiplied by an arbitrary bounded matrix sequence?

Since $Q_n$ is bounded in norm but its matrix size increases it seems natural to think that most of its entries must get very small as $n\to\infty$, but I believe this must happen at a rate faster than $1/\sqrt{n}$ in order to kill the growth rate $\sqrt{n}$ of $\left\lVert \Sigma_n \right\rVert_F$. But how can I show this rigorously?

Thanks a lot for your help.

  • $\begingroup$ The use of $n$ in your first and third paragraphs is a little unusual: is $Q_1$ a $ 1 \times 1$ matrix, and $Q_2$ a $2 \times 2$ matrix, etc.? (The problem I'm having is the phrase "a sequence of $n \times n$ positive matrices," which suggests that they're all of the same "type", like "a sequence of invertible matrices" --- the adjective "$n \times n$" seems to apply to all of them. Typically, this would mean that $n$ is a free (but fixed) parameter of the claim, but it's also a subscript on your matrices --- kind of a combination of a free and a bound variable. $\endgroup$ Apr 13, 2020 at 12:03
  • $\begingroup$ As for your question (assuming that the matrix sizes are indeed growing), I see no contradiction here. If, for instance, each $Q_n$ matrix were $I_n$, the identity, then the one sequence converging and the other diverging would be surprising, indeed, contradictory. But they cannot all be identity matrices, because that sequence isn't bounded. (In fact, it exactly provides the $\sqrt{n}$ in your first estimate.) Is this any more peculiar than that $\frac1n \to 0$, while $\sum \frac1n = \infty$, but when you multiply that divergent series by itself (term by term), the series actually converges? $\endgroup$ Apr 13, 2020 at 12:07
  • $\begingroup$ @JohnHughes Sorry for the confusion. Yes the matrix size is indeed growing with $n$. $\endgroup$
    – Alphie
    Apr 13, 2020 at 13:17
  • $\begingroup$ @JohnHughes I like your analogy. Still I don't get the intuition for the matrix case. $\endgroup$
    – Alphie
    Apr 13, 2020 at 13:27
  • $\begingroup$ @JohnHughes Since $Q_n$ is bounded but its matrix size increases it seems natural to think that most of its entries must get very small as $n\to\infty$, but this must happen at a rate faster then $1/\sqrt{n}$ to kill $\sqrt{n}\lambda^{1}_n$ no? $\endgroup$
    – Alphie
    Apr 13, 2020 at 13:34

1 Answer 1


Let $$\Sigma_n=\frac{1}{n^{\frac{1}{3}}}I_n$$ Then $$|\Sigma_n|_F=\sqrt{\Sigma \Sigma (a_{mn})^2}$$ $$=\sqrt{n \frac{1}{n^{2/3}}}$$ $$=n^{\frac{1}{6}}$$ although $\lambda_1=\frac{1}{n^{\frac{1}{3}}}$

  • $\begingroup$ Right but that's exactly my point: It is possible to have $\left \lvert \lvert \Sigma_n\right \rvert\rvert_F \to \infty$ but $\lambda_n^{1} \to 0$. $\endgroup$
    – Alphie
    Apr 15, 2020 at 14:52
  • $\begingroup$ My question is why is it the case that when such a matrix is multiplied by an arbitrary bounded sequence $Q_n$ of growing $n\times n$ matrices we have $\left \lvert \lvert \Sigma_n Q_n \right \rvert \rvert_F \to 0$ ? $\endgroup$
    – Alphie
    Apr 15, 2020 at 14:55
  • $\begingroup$ There are limits all over the place of the form $0 \times \infty$ that go to $0$. The restrictions of the principal eigenvalue going to zero and the bounded second matrix ensure this. $\endgroup$
    – Paul
    Apr 15, 2020 at 17:30
  • $\begingroup$ True. But in this case I feel like the limit is more of the form $M \times \infty$ where $M$ is the contant such that $\left \lvert \lvert Q_n \right \rvert \rvert_F \leq M$ for all $n$. Can I show that $\left \lvert \lvert Q_n \right \rvert \rvert_F \to 0$? $\endgroup$
    – Alphie
    Apr 16, 2020 at 8:32

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