How to prove that a nilpotent matrix is not invertible? Let $B_{n\times n}$ be a nilpotent matrix. How should I go around proving that $B$ is not invertible?
I was thinking this: if $B*B = 0$ , then if I rank $B$ to echelon form, I will always have two or more rows that are the same, and then because $B$ is square, it is not possible to find another matrix $(B^{-1})$ that will transform $B$ to $I_{n}$... But I don't know how to write it formally.
 A: Hint: If $AB = I$, then we must also have $A^kB^k = I$ for any $k$.
Or, if you prefer:
Hint: Note that $\det(B^k) = \det(B)^k$.

I don't think your argument is correct.  For example,
$$
B = \pmatrix{
0&1\\0&0
}
$$
is nilpotent, but doesn't have "two or more rows that are the same".
A: Usually, nilpotent means that $B^m=0$ for some $m>1 $, not necessarily  $2$.
A direct way to see that $B $ is singular is
$$
0=\det (B^m)=(\det (B))^m,
$$
so $\det (B)=0$.
Another way, without using determinants: if $B $ were invertible, then $$B=(B^{-1})^{m-1}\,B^m=0, $$ a contradiction.
A: Suppose $B^{-1}$ exists and $B^k=0$. Then 
$$
I = B^{-1} B=(B^{-1})^2 B^2=\ldots=(B^{-1})^k B^k = (B^{-1})^k 0 = 0
$$
which is a contradiction.
A: The eigenvalues of a nilpotent matrix is $0$ hence the determinant is also zero, implies it is not invertible.
A: Lemma: Let $A$ be a square matrix. If $AX=0$ and $A$ is invertible, then $X=0$.
Proof. We prove by minimum counterexample. Assume to the contrary that there exists an invertible matrix $A$ such that there exists $S \subset \mathbb{N}$ such that if $k \in S$, $A^k = 0$. Let $k = \min(S)$. If $k = 1$, then $A^1 = A =0$, which is not possible since $A$ is invertible and $0$ is clearly not; so $k > 1$. Hence, for all $1 \leq i < k$, $A^i \ne 0$.
Since $A^k = A A^{k-1} = 0$ and $A$ is invertible, by the above Lemma, $A^{k-1} = 0$, which is a contradiction.    
A: Nilpotent matrices must have strictly positive nullity, thus they are not invertible because they are not injective.
