A question on Rank of matrices Let $X_n=X+Y_n$ define a sequence of $m\times m$ random matrices. Let $Y_n$ be such that it converges in probability to a zero matrix. Then can it be that $Rank (X_n)$ for large enough $n$ be bigger than $Rank (X)$?
Is it possible to have $Rank(X)>0$ and come up with a scenario?
 A: Yes, simply choose $X$ to be the zero matrix and $Y_n$ a sequence of (not necessarily random) nonzero matrices that converge to zero.
Edit: You can take any arbitrary $X$ that does not have full rank. Let $Y$ be an invertible matrix. Now note that $\det(tX + Y)$ is a continuous map from $\mathbb{R}$ to $\mathbb{R}$ that converges to $\det(Y) \ne 0$ for $t \to 0$, i.e. the matrix $\det(tX + Y)$ is invertible for all $t$ with $|t| < \varepsilon$. But this means $\det(X + tY) = t^{-n} \det(tX + Y)$ is nonzero for $0 < t < \varepsilon$, so you can choose the sequence $Y_n = \frac{\varepsilon}{n}Y$.
I assume that a more careful exaimation should reveal that the set of matrices $Y$ so that $X + Y$ does not have full rank is a null set for any fixed matrix $X$.
A: Imagine $$X=\begin{pmatrix}1&0\\0& 0\end{pmatrix},$$ which has rank one. Now
$$Y_n=\begin{pmatrix}a^{11}_n&a^{12}_n\\a^{21}_n& a^{22}_n\end{pmatrix}.$$ Thus $$Y_n=\begin{pmatrix}1+a^{11}_n&a^{12}_n\\a^{21}_n& a^{22}_n\end{pmatrix}.$$ Can $X_n$ have rank two? It depends on the value of $$a^{22}_n(1+a^{11}_n)-a^{21}_na^{12}_n.$$ Is there any reason for being $0?$ Absolutely not, even if $Y_n\to 0.$ So we can have $\rm{rank}(X_n)>\rm{rank} (X)$ for $n$ big enough. Even more, for all $n.$
Note that if $\rm{rank}(X)=m$ then $\rm{rank}(X_n)\le \rm{rank}(X).$
