Finding the rank of a matrix with zero and nonzero eigenvalues I have a square matrix that has zero and nonzero eigenvalues. I need to find the rank of this matrix. Can I say that the rank is: the dimension of the matrix - the number of nonzero eigenvalues counting each once (ignoring their algebraic multiplicity)? 
If so, how can I support this expression with theorems? 
I ask this question because I don't want to deal with the nullity theorem.
Thanks for the help. 
 A: This is more or less backwards. Instead, the rank is the dimension of the matrix minus the number of zero eigenvalues counting their geometric multiplicity, which is essentially the statement of the rank nullity theorem. 
It's easy to construct a counter example to your statement. For example, your approach would conclude that the rank of the matrix 
$$
\begin{pmatrix}
1 & 0 \\
0 & 2
\end{pmatrix}
$$
is equal to $2-2=0$, since it has two distinct eigenvalues. This is of course not the case. 
You also can't just count geometric or algebraic multiplicity of nonzero eigenvalues to get the rank. For example
$$
\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}
$$
has no nonzero eigenvalues but has rank $1$. 
I'd highly suggest getting comfortable with the rank-nullity theorem if you are going to be doing a lot of linear algebra moving forward. Its a rather powerful tool and understanding it is important for understanding matrices in general.
A: Not true. Take, for instance,$$A=\begin{bmatrix}0&0&0\\0&1&0\\0&0&-1\end{bmatrix}.$$Then the rank of $A$ is $2$, not $1$.
A: No. Take for example the $ 3 \times 3$ - matrix $A= diag(0,1,1)$. We have that $rank(A)=2$ but  the number of nonzero eigenvalues counting each once (ignoring their algebraic multiplicity) $=1$.
