Matrix rank and probability limit

Question

If I have a random matrix $M$ (dimension $l\times l$) whose elements are stochastic and depend on $n$ such that for all $n$, the rank of $M$ is less than or equal to $k$, can I infer that $$ \text{plim}_{n\to\infty}M $$ has rank less than or equal to $k$? Here, we assume that the probability limit above exists and $l$ and $k$ do not depend on $n$.

The result appears as an implicit step in a proof in my textbook. I know the Continuous Mapping Theorem but rank isn't a continuous map of a matrix so I'm stuck. Thanks for your help.

Answer 1

This follows from two facts:

1. The set of $l \times l$ matrices with rank $\leq k$ is closed.

This is true since, for $k<l$, the matrices with rank $\leq k$ are precisely those for which every $(k+1)\times (k+1)$ submatrix has determinant 0. (Now use the fact that the determinant is continuous, so its preimage of $\{0\}$ is closed, and the intersection of closed sets is closed.)

2. Given a closed subset $C$ of a metric space $E$ and a sequence of random elements $X_i$ of $E$, if $X_i$ converges in probability to a random variable $X$, and if $P(X_i \in C)=1$ for all $i$, then $P(X \in C)=1$.

This is true since convergence in probability implies a.s.-convergence on a subsequence.