Prove that if $A^2=0$ then $A$ is not invertible Let $A$ be $n\times n$ matrix. Prove that if $A^2=\mathbf{0}$ then $A$ is not invertible.
 A: If $A$ is invertible then $I=AB$ for some $B$. Then $A=AI=A^2B=0$, and so $A=0$, but $0$ is not invertible.
A: If $A,B \in \mathbb{R}^{n \times n}$, then $\text{rank}(A) + \text{rank}(B) - n \leq \text{rank}(AB) \leq \text{min}(\text{rank}(A),\text{rank}(B))$.
Here $A=B$ and $A^2 =0$. Hence $\text{rank}(A^2) = 0$.
Hence if $\text{rank}(A) = k$, then we have $$k+k - n\leq 0 \leq k$$
From which we get that $0 \leq k \leq \frac{n}{2}$.
So this implies not only is the determinant $0$ and there is no inverse but also there cannot be more than $\frac{n}{2}$ linearly independent rows/columns of the matrix $A$.
A: With no determinant. 
Assume that $A$ is invertible on the right. This means that there exists $B$ such that $AB=I$, hence $A^2B=A$. Since $A^2=0$, one knows that $A^2B=0$, hence $A=0$. But the matrix $0$ is not invertible on the right, hence the hypothesis was false. This proves that $A$ is not invertible on the right. 
Likewise if one assumes that $A$ is invertible on the left, that is, that there exists $B$ such that $BA=I$.
As explained in a comment, in the infinite dimensional case, the fact that $A$ is not injective (as proved by Andres) prevents $A$ to be invertible on the left but not on the right.
A: This is simply a special case of the fact that, in every ring, a unit is not a zero-divisor, $\:$ since $\rm\ A' A = 1,\ AB = 0\ \Rightarrow\ B = (A'A)B = A'(AB) = 0\:.\: $ The problem is the special case $\rm\ B = A\:.$ 
Informally: a ring collapses to the zero ring if one adjoins an inverse of $0$ (or a zero-divisor).
A: I think that for a problem like this, using determinants (or invoking the rank-nullity theorem) ends up hiding what really goes on.
Instead, I suggest that you simply argue about this directly: $A^2=0$ means that $A^2v=0$ for any $v$. Then $A(Av)=0$ for any $v$. 
If for some $v$ we have $w=Av\ne 0$, then $A$ is not injective, because $$Aw=A(Av)=A^2v=0,$$ and we are done. 
Else, $Av=0$ for all $v$, and then $A=0$, which of course is not invertible.
A: $0=\det (A^2) = (\det A)^2$. Hence $\det A=0$.
A: Well I've heard that the more ways you can prove something, the merrier. :) So here's a sketch of the proof that immediately came to mind, although it may not be as snappy as some of the other good ones here:
Let's prove the contrapositive, that is if $A$ is invertible then $A^2 \neq 0$.
If $A$ is invertible then we can write it as a product of elementary matrices,
$$A = E_n...E_1I$$
Then $A^2$ can be written as
$$AA = (E_n...E_1I)(E_n...E_1I) = (E_n...E_1 E_n...E_1)I$$
which is a sequence of elementary row operations on the identity matrix. But this will never produce the zero matrix $0$. QED.
A: Assume there is an inverse denoted A-1. Then you have AA = 0 and then multiply it by your inverse A-1 you have AAA-1 = 0A-1 = A = 0.
Now given the definition of 0 :
Given a Matrix M and the0vector M0=0=0M.
Therefore by definition of0it cannot have an inverse.
