Vector Space as the set of solutions of matrix equation $AX=O$ One of my professor's lecture notes on Vector Spaces start by the following lines:-
We have seen that if $det(A)$ = 0, then system $AX=O$ has infinite number of solutions. We shall now see that in this case, the set of solutions has a structure called vector space.
My doubt is in what sense do the set of an infinite number of the solution of equation $AX = O$ (given |A|=0) is actually a structure of Vector Space? How does the term Vector Space come into picture?
 A: What your professor means is that the set of all $X$'s that satisfy $AX=0$ satisfies the properties of a vector space. You have to check all the ten axioms of a vector space. Here is only a few:
It has an identity under vector addition. Namely, $X=(0,0,\cdots,0)^t$.
It's closed under addition because $A(X+Y)=AX+AY=0$ and if $X,Y$ are in the set, so is $X+Y$
It's closed under scalar multiplication because $A(\lambda X)=\lambda AX=0$.
And you can check all other axioms one by one. Another way to avoid checking many of the axioms, is that if you already have shown that $\mathbb{F}^n$ is a vector space for any field $F$. Then, since our set is a subset of that, associativity and many other properties are inherited from the superset naturally. And we only need to show that $\alpha X + \beta Y$ is in the set which follows from what we proved about being closed under addition and scalar multiplication.
A: It's simply the fact that linear combinations of solutions are again a solution: if $AX=0$ and $AY=0$, then 
$$
A(\alpha X+\beta Y)=\alpha AX+\beta AY=0.
$$
Thus, the set $\{X:\ AX=0\}$ is a vector space with the natural vector operations. 
A: You must look into the actual definition of a vector space... In this this case, since the set of solutions of $AX=0$ is a subset of $\mathbb{R}^n$, you just need to check that it is closed with respect to the sum of vectors and to the multiplication by scalars. Take $u,v \in \mathbb{R}^n$ such that $Au = Av = 0$ (basically any two solutions of $AX=0$) and $\alpha \in \mathbb{R}$. If you prove that $u+v$ and $\alpha u$ are also solutions of $AX=0$, you have your result. This is very straightforward to check:
$$ A(u+v)= Au + Av = 0 + 0 = 0$$
$$A(\alpha u) = \alpha (Au) = \alpha \cdot 0 = 0$$
