What does $F^2$ mean in terms of vector space/subspaces "I am learning linear algebra and I am not sure what is meant by the $F^2$ notation.  It was stated previously that $F = \mathbb{Z}_3$ with elements $[0]$, $[1]$, and $[2]$. Does $F^2$ truly square $F$? Do all the elements get squared as well? (i.e. would the elements be $[0]$, $[1]$, and $[4]$?)
 A: Nope, I think that $F^2$ means $F\times F$, just as $\mathbb R^3$ means $\mathbb{R\times R\times R}$.
(By the way, in $\mathbb{Z}_3$, $[4]=[1]$)
A: In a sense it's squaring, yes. But it's more abstract than just squaring the elements. It's squaring the set itself.
Given two sets $X, Y$, the product of $X$ and $Y$, commonly denoted $X\times Y$, is the set of all pairs $(x, y)$ with $x\in X, y\in Y$. Some times, $X$ and $Y$ will be the same space, and you end up with $X\times X$, which is often shortened to $X^2$.
(You may be familiar with the plane, where each point has a coordinate pair $(a, b)$ where $a$ and $b$ are real numbers. This plane is often called $\Bbb R^2$, because it's exactly this contsruction applied to the set $\Bbb R$ of real numbers.)
In your case, we're looking at $(\Bbb Z_3)^2$. The elements are all possible pairs $([a],[b])$, where $[a], [b]\in \Bbb Z_3$. There are therefore nine elements (note that $9 = 3^2$, which is not a coincidence, and I believe a major reason that this construction is called a "product" in the first place: For finite sets, the number of elements in a product set is the product of the number of elements in the "factor" sets). Those elements are
$$
([0],[0]),\quad([0],[1]), \quad([0],[2])\\
([1],[0]),\quad([1],[1]), \quad([1],[2])\\
([2],[0]),\quad([2],[1]), \quad([2],[2])
$$
If you are familiar with vectors in the real plane, these work exactly the same way. For instance, you add two of them by adding the first components together and adding the second components together. You scale by an element in $\Bbb Z_3$ by scaling each component.
