Why can you diagonalize a matrix, if a certain eigenvector gets scaled by 3, and there are 2 other eigenvectors that also get scaled by 3. Imagine you have a matrix $A$ with eigenvalues $ (1, 3, 3)$.
The algebraic multiplicity is of the eigenvalue $3$ is $2$. So this translates to that there are two lines in vector space $V$, where along this line all vectors get scaled by $3$.
The geometric multiplicity is the number of independent eigenvectors associated with an eigenvalue if I understand correctly.
Then if GM = AM for all eigenvalues, then the matrix is diagonalizable and vice versa.
How does this make sense, can't you always choose the basis all the eigenvectors that get scaled by $3$ and use that for the basis as the eigenspace? Wouldn't this always result in GM = AM?
What does this exactly have to do if a matrix is diagonalizable?
 A: Your definition of algebraic multiplicity is incorrect.  In fact, if I understand what you're saying correctly, it's basically equivalent to your (correct) definition of geometric multiplicity.  That would explain why you concluded that the two are always equal to each other.
The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial of the matrix.  For example, the matrix
$$
M = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{bmatrix}
$$
has characteristic polynomial $(\lambda - 1) (\lambda-3)^2$, and so 3 is an eigenvalue of $M$ with algebraic multiplicity ${AM} = 2$.
However, the dimension of the subspace of vectors $\vec{v}$ such that $M\vec{v} = 3 \vec{v}$ (the geometric multiplicity) is only $GM = 1$.  We can see this by noting that
$$
M - 3 I =  \begin{bmatrix} -2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}
$$
and the nullspace of this matrix is one-dimensional, consisting of multiples of $[0 \, 1 \, 0]^T$.
A: If you look at, for example, $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, the matrix for a right shear.
It has an AM of 2 for the eigenvalue 1, yet it doesn't map all vectors to themselves, because while the only singular matrix of the form $\lambda I-A$ is $\lambda=1$, it results in the matrix $I-A=\begin{pmatrix}-1&0\\0&0\end{pmatrix}$ which only has a 1D solution space, spanned by $v_1=\begin{pmatrix}1\\0\end{pmatrix}$.
What happens to other vectors then?
Well, let's look at $v_2=\begin{pmatrix}0\\1\end{pmatrix}$.
$$Av_2=v_2+v_1$$
The reason no other eigenvectors exist is that $A(av_1+bv_2)=(a+b)v_1+bv_2$, so for it to be an eigenvector, it must have eigenvalue 1, which implies it is $v_1$.
