# If $A^2$ is invertible, then $A$ is also invertible?

True or False: If $A^2$ is invertible, then $A$ is also invertible.

($A$ is a matrix here.)

The answer is true. I was trying to come up with an example that makes this false.

But I couldn't. Could anybody help me prove this?

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Hint: Suppose $B$ is the inverse of $A^2$. That is, let $B$ be the matrix such that $(A^2)\cdot B=I$ where $I$ is the identity matrix. Note that matrix multiplication is associative, so $$I=(A^2)\cdot B=(A\cdot A)\cdot B=A\cdot(A\cdot B).$$ Do you see the inverse to the matrix $A$?

I am implicitly using the fact that (for square matrices) a one-sided inverse, for either side, will also necessarily be a two-sided inverse. Here is the math.SE thread about this fact.

• (I posted a comment earlier, but then I decided you can repeat the same argument with $B(A^2)$ instead of $(A^2)B$ to find an inverse on the other side. It's then easy to prove that if a matrix has both a left inverse and right inverse, they have to be equal. On the other hand, it is somewhat nontrivial to prove that a left inverse will automatically be a right inverse. Sorry to cause an edit!) Apr 23, 2013 at 23:50

Since $A^{2}$ is invertable so $\det(A^{2})\ne 0$ . On the other hand $\det(A^{2})=\det(A)\cdot\det(A)$ and so $\det(A)\ne 0$ so $A$ is invertible too.

• This is a correct solution, but it makes unnecessary use of the determinant concept. The proof given by Zev Chonoles shows the true nature of the result. It is a purely algebraic fact that is true in many algebraic systems, not just matrices. Apr 23, 2013 at 23:32
• A LaTeX tip: the determinant is correctly written with upright letters, in the same way as the sine function and logarithm function are ($\sin\theta$ produces $\sin\theta$, $\log x$ produces $\log x$, and $\det A$ produces $\det A$). Apr 23, 2013 at 23:46

Hint: what happens if you multiply $A$ by $A(A^2)^{-1}$?

If $A^2$ is invertible, there exists $B$ such that $A^2B=I$, where $I$ is the Identity... Therefore $A(AB)=I$, and then $A$ is invertible; its inverse being $AB$.

• Good ! Very clear!
– mick
May 19, 2013 at 19:03

If matrix $A$ is a transformation which is not invertible, then applying $A$ twice to make $AA = A^2$ also cannot be invertible.

If $A$ is not invertible it means that $y = Ax$ applies a transformation to a space of vectors $x$ which irretrievably destroys information is needed to map the resulting vectors $y$ back to the original values $x$. $Ax$ is a "trap door": a "one way function".

There is nothing which can multiply $Ax$ to recover the lost information, let alone another $A$!

So it is impossible for $AA$ to be invertible without $A$ being invertible.

Think the other way:

If $A$ is not invertible, could $A^2$ be invertible ?

Invertible means bijective which is equivalent to injective or surjective.

If $A$ is not injective, could $A^2$ be injective ?

No, it means that there exist $x_1,x_2$ such that $Ax_1=Ax_2$. Just multiply it by $A$, then there exists $x_1,x_2$ such that $A^2 x_1=A^2x_2$ and $A^2$ is not injective. $A^2$ is not invertible.

In term of Kernel, $A$ injective means $Ker(A) \neq [o]$. As we clearly have $Ker(A) \subset Ker(BA)$ for all B, taking $B =A$, gives $Ker(A^2)\neq [o]$. $A^2$ is not injective. $A^2$ is not invertible.

If $A$ is not surjective, could $A^2$ be surjective ?

It means that there exist a vector y such that there is no x that solve $Ax=y$. So there is no x that solve $AAx=z$ with $z=Ay$. $A^2$ is not surjective. $A^2$ is not invertible.

We clearly have $Im(AB) \subset Im(A)$, so for $B=A$ and with that $A^2$ surjective means $Im(A^2) = \Re^n$. Then by inclusion $Im(A) = \Re^n$. Then A is surjective. $A^2$ is not invertible.

There is faster solutions but i thinks you always should keep in mind the equivalence between bijective, surjective and injective (in finite dimension).