I read somewhere that this particular vector space,$$M(A,\alpha):=\{v\in F^n \mid (A-\alpha I_n)^nv=0\},$$ where $F$ is the splitting field for the characteristic polynomial of $A$ and $F^n$ is the vector space containing all column vectors of length $n$, has dimension equivalent to the algebraic multiplicity of $\alpha$ as an eigenvalue of $A$.
Can someone explain how this thing happened? I'm quite a newbie in stuffs like this. Thank you very much for those who would gladly help.
$M(A,\alpha)$ is called the generalised eigenspace for the eigenvalue $\alpha$. It is in fact equal to $\text{Ker} (A-\alpha\text{Id})^d$ where $d$ is the algebraic multiplicity of $\alpha$.
One reason why its dimension is $d$ is that $F^n$ can be proved to be the direct sum of all generalised eigenspaces, and that each of them has dimension at most equal to the corresponding algebraic multiplicity. Therefore all of these inequalities are equalities for dimensional reasons.
Another way to see it is that in these settings, $A$ has a Jordan normal form, i.e. is similar to a triangular matrix with the eigenvalues on the diagonal, ones just above the diagonal, and zeroes elsewhere. From there the result is pretty easy.
Digression.
If you start from a vector $v$ and apply $A-\alpha\text{Id}$ once, then if you get $0$ it means that $v$ is an $\alpha$-eigenvector. But if you don't get $0$ you can iterate this process; if you get $0$ at the second iteration it means that in restriction to the $2$-dimensional space $\left<v, (A-\alpha\text{Id})v\right>$, with these basis vectors, the matrix of $A$ is $$\left(\matrix{\alpha & 0\\1&\alpha}\right)$$ This is the baby case of a Jordan block.
Note that I'm not saying that you will get $0$ eventually no matter which vector you started from. You will only if you started from a vector that does lie in $M(A,\alpha)$.