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?

Solution 1:

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.

Solution 2:

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

Solution 3:

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.

Solution 4:

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 $AB$.

Solution 5:

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.