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Let $X_n \in \mathbb R^{n \times k}$ with $\mathop{\mathrm{rk}}\left(X_n\right) = k$.

Does $\lim_{n \to \infty} \left(X_n^\top X_n \right)^{-1} = 0_{\mathbb R^{k \times k}}$ hold if and only if $\lim_{n \to \infty}\mathop{\mathrm{tr}}\left[\left(X_n^\top X_n \right)^{-1} \right] = 0$?

I know that $\lim_{n \to \infty}\mathop{\mathrm{tr}}\left[\left(X_n^\top X_n \right)^{-1} \right] = 0$ implies that the positive square root of $\left(X_n^\top X_n \right)^{-1}$ vanishes, but I'm not sure if I'm on the right track with that.

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4 Answers 4

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You certainly know that if $A_n:=\left(X_n^\top X_n \right)^{-1}\to0_{\mathbb R^{k \times k}}$, then $\mathop{\mathrm{tr}}A_n\to0$ (by continuity of the trace map). Your question is then only about the converse.

Since each $A_n$ is symmetric positive, it can be written $B_n^\top B_n$ for some square matrix $B_n$, hence your question can be reworded:

Does $\|B_n\|_{\text{Frobenius}}^2:=\mathop{\mathrm{tr}}\left(B_n^\top B_n\right)\to0$ imply $B_n^TB_n\to0_{\mathbb R^{k \times k}}$?

You wrote in comment you "knew $\|B_n\|_{\text{Frobenius}}\to0\iff B_n\to0$, but didn't know that this implies $B_n^TB_n\to0$".

Well, it does: $$B_n\to0\implies B_n^TB_n\to0,$$ since transposition and matrix product (on fixed matrix spaces) are continuous.

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From another perspective, for a positive definite matrix $P$, its matrix trace $\operatorname{tr}(P)$ is identical to its trace norm (a.k.a. Schatten $1$-norm or nuclear norm) $\|P\|_{\text{tr}}=\sum_i\sigma_i(P)$.

Since your $X_n$ has full column rank, $X_n^TX_n$ is a $k\times k$ positive definite matrix and so is $P_n=(X_n^TX_n)^{-1}$. So, you are essentially asking whether $P_n\to0$ if and only if $\|P_n\|_\text{tr}\to0$. Since all norms are equivalent in a finite-dimensional vector space, the answer to your question is positive.

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If $A_n$ is positive definite then $\mathrm{trace} (A_n)=\|A_n^{1/2}\|^2\to 0$ implies $A_n^{1/2}\to 0$, therefore $A_n\to 0$. Apply this to $A_n=(X_n^TX_n)^{-1}.$

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This is really a question about $k\times k$ positive semi-definite matrices $A\succeq 0$; the inverse you described falls in that class. I'll assume that the limit you are working with is in terms of Frobenius norm, correct me if that is not the case. However, as long as the chosen norm induces the same topology, results should match.

For PSD matrices, the trace is the sum of eigenvalues $\mathrm{tr}(A)=\sum_{i=1}^k\lambda_{(k)}$, and the (squared) Frobenius norm is the sum of squared eigenvalues $\|A\|_F^2=\sum_{i=1}^k\lambda_{(k)}^2$.

All eigenvalues are positive, so for fixed trace $\mathrm{tr}(A)=C$, the Frobenius norm is minimized when $\lambda_{(1)}=...=\lambda_{(k)}=\frac{C}{k}$ and $\|A\|_F=\frac{C}{\sqrt{k}}$. Likewise, the maximum occurs when $\lambda_{(k)}=C$, $\lambda_{(k-1)}, ..., \lambda_{(1)}=0$, and $\|A\|_F=C$. Then we have the general inequality $\|A\|_F\leq \mathrm{tr}(A)\leq \sqrt{k}\cdot\|A\|_F$. From this interleaving, it follows that both quantities must go to $0$ together, provided $k$ is bounded as $n\to\infty$.

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    $\begingroup$ With $\lim_{n \to \infty} A_n = 0$ I mean $\lim_{n \to \infty} \left(A_n\right)_{ij} = 0$ for all $i, j = 1, \ldots, k$; say, w.r.t. the Euclidean norm. $\endgroup$
    – statmerkur
    Mar 24, 2023 at 9:36
  • $\begingroup$ a) in a finite dimensional spaces, all norms are equivalent (not so trivial theorem, proved in undergraduate classes) b) For matrices the good norm comes from the euclidean structure given by the scalar product $\langle A,B\rangle =\mathrm{trace }A^TB$ called the Fobenius norm and which in fact the one that you are using. $\endgroup$ Mar 26, 2023 at 14:03

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