Proof about diagonal matrices Suppose that $A\in M_{n\times n}(\mathbb{R})$ such that their eigenvalues are $\{\lambda_1,\cdots, \lambda_n\}$, i.e. $\sigma(A)=\{\lambda_1,\cdots,\lambda_n\}$, then if the geometric multiplicity $mg_A(\lambda_i)$ is the same arithmetic multiplicity $ma_A(\lambda)$, we have that $A$ can be diagonal
 A: In the Jordan form of the Matrix $A$, for each $\lambda_i$, the number of different Jordan blocks corresponding to this eigenvalue is equal to geometric multiplicity of $\lambda_i$. The geometric multiplicity of $\lambda_i$ is indeed $\dim\, \operatorname{null} (A-\lambda_i I)$. 
By arithmetic multiplicity of $\lambda_i$, I guess you mean algebraic multiplicity of $\lambda_i$ which is defined as the multiplicity of $\lambda_i$ in zeros of characteristic polynomial of $A$. It is also equal to sum of the orders of all Jordan block corresponding to $\lambda_i$.
In this sense, the algebraic multiplicity of $\lambda_i$ is not bigger than or equal to geometric multiplicity of $\lambda_i$.  
However in case of equality, we know that if the algebraic multiplicity of $\lambda_i$ is $k$, then one can find $k$ linearly independent eigenvectors of  $\lambda_i$ which constitute a basis for $(A-\lambda_i I)$. As a result one can find $n$ linearly independent eigenvectors for $A$ which means that it can be diagonalized. 
To look at it differently, if number of Jordan blocks of $\lambda_i$ is equal to the sum of orders of these block, then each block should have the order 1. This means than Jordan canonical form of $A$ is diagonal.
The idea behind this is that for equal algebraic-geometric multiplicity you can find $n$ linearly independent eigenvectors as a base for $\mathbb R^n$ which can be used later for diagonalizing $A$.
