Show that the eigenvalues of this positive semi-definite matrix converge Let $(H_n)$ and $(\Lambda_n)$ be sequences of $n\times k$ matrices respectively (growing number of rows), where $k\in\mathbb{N}$ is fixed. We make the following two assumptions:

*

*$H'_nH_n \to I_k$ as $n\to \infty$.


*$\Lambda'_n\Lambda_n \to \Sigma$ as $n\to \infty$, where $\Sigma$ is  a positive definite matrix with distinct eigenvalues.
Here convergence is with respect to the Frobenius norm, which I denote $||\cdot||$.
I want to show that the largest $k$ eigenvalues of the matrix $H_n \Lambda'_n \Lambda_n H'_n$ converge to those of $\Sigma$ (the rank of $H_n \Lambda'_n \Lambda_n H'_n$ is at most $k$, so all other eigenvalues must be zero).
Any ideas on how to proceed is greatly appreciated.
 A: This one is surprisingly easy, as long as the $k$ leading eigenvalues are non-zero.
It is well known that the eigenvalues of an $n \times n$ matrix depend continuously on the entries of that matrix. With that said, we have
$$
\vec \lambda^{\downarrow k}(H_n \Lambda'_n \Lambda_n H_n') = \vec \lambda^{\downarrow}(\Lambda_n' \Lambda_nH_n'H_n)
$$
as a consequence of the fact that $H_n \Lambda'_n \Lambda_n H_n'$ is positive semidefinite and the fact that $AB$ and $BA$ have the same non-zero eigenvalues (for any matrices $AB$ such that both $AB$ and $BA$ are square).
It follows that
$$
\lim_{n \to \infty}\vec \lambda^{\downarrow k}(H_n \Lambda'_n \Lambda_n H_n') = \\
\lim_{n \to \infty}\vec \lambda^{\downarrow}(\Lambda_n' \Lambda_nH_n'H_n) = \\
\vec \lambda^{\downarrow}\left(\lim_{n \to \infty} \Lambda_n'\Lambda_n H_n' H_n \right) = \\
\vec \lambda^{\downarrow}\left([\lim_{n \to \infty} \Lambda_n'\Lambda_n] [\lim_{n \to \infty}H_n' H_n] \right) = \\
\vec \lambda^{\downarrow}(\Sigma I_k) = \vec \lambda^{\downarrow}(\Sigma),
$$
which is what we wanted.
