Consider an $n\times m$ matrix with i.i.d. entries each having zero mean and variance $1/n$.
Let $Y = X^TX$.
By the strong law of large numbers, we know that the $(i,j)$ entry of $Y$ goes almost surely to $\delta_{i,j}$ (the Kronecker delta) for all $1\leq i,j\leq m$ as $n\to\infty$.
Now, in some book the authors claim that the matrix $Y$ goes almost surely to the identity matrix when $m$ is fixed.
My first question: What does it means that some matrix goes almost surely (or even in probability) to another matrix? Does it means that the union of all the events of deviations of each entry goes to zero (in case of convergence in probability)?
My second question: Actually this question depends on the answer to the previous question, but, assuming that the notion of matrix convergence is as I "described" in my previous question, then I wonder why $m$ is must be fixed?
To answer 1: Almost-sure convergence means exactly what it should: $Y(=Y_n)$ converges almost surely to some (random) matrix $Z$ if $P(\lim_{n\rightarrow\infty}Y_n=Z)=1$.
Now, what does this mean? Well, it depends on how you define the limit of a sequence of matrices; typically, this is done with respect to some norm on the set of such matrices. While we are sitting on the space of $m\times m$ matrices, all norms are equivalent (since the space is finite-dimensional), and so you could be talking about, say, treating both as $m^2$-long vectors and using the Euclidean norm; it could be the operator norm; etc. See, for instance, Wikipedia.
For 2: note that if $m$ is allowed to change with $n$, then the size of your matrix $Y_n$ actually changes with $n$ -- and it is terribly unlikely that you can discuss convergence of a sequence of matrices that have different sizes!
It is possible that you might be able to prove that if $m=m(n)$ and $Z_n$ is a sequence of matrices with $Z_n$ being $m(n)\times m(n)$, then $\|Y_n-Z_n\|\rightarrow0$ almost surely.