# 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.

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Please don't order the group around. If you have a question, ask a question. –  Arturo Magidin Feb 15 '11 at 18:26
I suppose it would be too late to ask everybody not to post complete answers to homework questions... –  Qiaochu Yuan Feb 15 '11 at 19:28
Here $n$ should be positive, for otherwise the statement is false. –  Mariano Suárez-Alvarez Feb 15 '11 at 21:15
@Rasmus, as the only element in that ring is $0$, every element satisfies the condition that $A^2=0$, but every element is also invertible! –  Mariano Suárez-Alvarez Mar 12 '11 at 15:56
@Mariano: Right. I confused myself - sorry. Thank you for your explanation. –  Rasmus Mar 12 '11 at 16:10

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).

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With no determinant. Assume that $A$ is invertible on the right, thus there exists $B$ such that $AB=I$, hence $A^2B=A$; since $A^2=0$, $A^2B=0$, 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.

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Sorry, could you please explain how $A^2B=0$ makes you conclude $A=0$? –  wvxvw Nov 15 '14 at 17:01
@wvxvw A^2B=A and A^2B=0 hence A=... –  Did Nov 15 '14 at 17:28
Oh, thank you, I see now. –  wvxvw Nov 15 '14 at 17:43

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.

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+1 I like this proof most over the others, because it reinforces the fact that matrices are functions, which the algebraic manipulations hide somewhat. –  Uticensis Feb 15 '11 at 19:30
Note also that nothing in this proof requires A to act on a finite-dimensional vector space. Thus, unlike the argument involving either determinants or the rank-nullity theorem, this argument generalizes to arbitrary vector spaces. –  Qiaochu Yuan Feb 15 '11 at 19:32
In the infinite dimensional case, the fact that $A$ is not injective prevents $A$ to be invertible on the left but not on the right. That is, there exists no $B$ such that $BA=I$ but there could exist some $B$ such that $AB=I$. –  Did Feb 16 '11 at 7:08
One could also note that a composition of injective maps is injective, and a composition of surjective maps is surjective (thus ruling out both forms of 1-sided invertibility as mentioned in Didier Piau's comment). –  Jonas Meyer Feb 16 '11 at 15:55

$0=\det (A^2) = (\det A)^2$. Hence $\det A=0$.

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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.

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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 $A$.

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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.

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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.

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