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!!
 A: If a $n\times n$ matrix has $n$ distinct eigenvalues, then it is automatically diagonalizable. Otherwise, compute the dimension of each eigenspace. The matrix is diagonalizabel if and only if the sum of these dimensions is $n$.
Concerning the sentence “There exists a matrix $B$ and a diagonal matrix $D$ such that: $D=B^{−1}AB$”, well… That's basically what being a diagonalizable matrix means.
A: The statement "matrix $A$ ($n×n$), is diagonalizable if: number of eigenvectors equals to number of eigenvalues" is not correct. 
We would say that a matrix $A$ ($n×n$), is diagonalizable if and only if the sum of the dimension of eigenspaces is equal to $n$, that is if and only if for any eigenvalues the algebraic multiplicity is equal to the geometric multiplicity.
When a matrix is diagonalizable, of course, by definition the diagonal form is similar to the original matrix. Note that similarity holds, more in general, also with the Jordan normal form when the matrix is not diagonalizable.
A: 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.

