Sufficient condition for a matrix to be diagonalizable and similar matrices my question is about diagonalizable matrices and similar matrices.
I have a trouble proving a matrix is diagonalizable. 
I know some options to do that:
Matrix $A$ $(n \times n)$, is diagonalizable if:


*

*Number of eigenvectors equals to number of eigenvalues.

*There exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^{-1}AB$. 


But i have a trouble to determine it according the second option, Do i really need to search if there exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^{-1}AB?$
I really sorry to ask an additional question here: If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
in general, given a matrix, how do i know if is a diagonalizable matrix? Are there some additional formulas to do that?
Thanks for help!!

First a comment
The wording Number of eigenvectors equals to number of eigenvalues... is confusing. If $A$ has a non zero eigenvector then $A$ has an infinite number of eigenvectors (providing you work in $\mathbb R$ or $\mathbb C$ for example). A proper wording would be $A$ has a basis of eigenvectors.
If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
The implication "If a matrix has a row of $0$'s" then "that matrix is diagonalizable" is not true.
The matrix
$$A=\begin{pmatrix}
0 & 0\\
1 & 0
\end{pmatrix}$$ is an example. The only eigenspace is $\mathbb F e_2$ were $e_2$ is the second vector of the canonical basis (and $\mathbb F$ the field of the vector space).
Some equivalent conditions for a matrix $A$ to be diagonalizable


*

*The sum of the dimensions of its eigenspaces is equal to the dimension $n$ of the space.

*$A$ is similar to a diagonal matrix.

*Its minimal polynomial is a product of distinct linear factors.

