Let's start with a simplified "universe" (his term, not mine), the integers from $0$ to $63$. There are $64$ total elements in this universe, and we see that $64 = 2^6$. In other words, for every key in our universe, I can represent it as a vector in base-$2$ with six coordinates. For example, $42$ can be represented as $(1, 0, 1, 0, 1, 0)$, and $13$ can be represented as $(0,0,1,0,1,1)$. Here I'm taking the leftmost entry to be the most significant digit. These representations look awfully familiar: concatenate the entries, and you have the binary representation of a number: $42_d = 101010_b$, $13_d = 001101_b$, etc.
Take a moment to remember our ultimate goal in creating a hashing function: we want to map any key to an integer between $0$ and the number of "buckets," $m$. He defines the hash function as follows: take the dot product of some randomly pre-chosen key $a$ and the key $k$ which you wish to hash, then take the result modulo $m$ (this is to ensure that the hash function returns a number between $0$ and $m-1$, inclusive). So if I'm trying to hash $13$ and my pre-chosen key $a$ was $42$, then I would first compute the dot product of their vector representations: $(1,0,1,0,1,0) \cdot (0,0,1,0,1,1) = 2$, then take the remainder when divided by $m$ ($m$ = 2 in this example, so our hash function returns $0 \equiv 2 \text{ mod } 2 $).
I'm not quite sure what you mean by this statement: "I also used to figure a suitable matrix out of this family and couldn't fine one," but I hope the above explanations helped!
Best Answer
Once the hash function has been chosen for a given key, the key ought to remain in that slot. The whole point of a class of universal hash functions is that the runtime for data retrieval is expected to be significantly less than the worst case in a basic hash using chaining. Specifically, it can be shown that the expected length of the linked list corresponding to an arbitrary key is equal to $1+\frac{n}{m},$ where $n$ is the number of keys and $m$ is the number of slots (cf. Cormen et. all.) Perhaps I don't fully understand your question though.