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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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  • $\begingroup$ asd213: When you receive answers that are helpful, we encourage users who ask questions to accept one that they found helpful. (You can only accept one answer per question). To accept an answer, simply click on the $\checkmark$ to the left of the answer you'd like to accept. Plus, you get two reputation points for each answer you accept. You can also upvote as many answers as you'd like! $\endgroup$
    – amWhy
    Commented Apr 25, 2013 at 3:08

6 Answers 6

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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.

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    $\begingroup$ (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!) $\endgroup$
    – Jason DeVito - on hiatus
    Commented Apr 23, 2013 at 23:50
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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.

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    $\begingroup$ 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. $\endgroup$
    – Ittay Weiss
    Commented Apr 23, 2013 at 23:32
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    $\begingroup$ 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 $\det A$). $\endgroup$
    – Zev Chonoles
    Commented Apr 23, 2013 at 23:46
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Hint: what happens if you multiply $A$ by $A(A^2)^{-1}$?

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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 $AB$.

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    $\begingroup$ Good ! Very clear! $\endgroup$
    – mick
    Commented May 19, 2013 at 19:03
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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.

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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).

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