Neat Questions! Invertible Power of a Matrix.

September 25, 2010

On a linear algebra exam, I was asked something like, “If $A$ is a real $n\times n$ matrix and $A^{2}$ is invertible, prove that $A$ is also invertible.”

Apparently, a number of students attempted the problem this way:

$A^{2} * (A^{2})^{-1} = A * (A * A^{-1}) * A^{-1} = A * A^{-1} = 1$

and concluded that $A$ had an inverse.  What’s wrong with this argument?  The astute reader will note that we’re actually assuming that $A^{-1}$ exists to prove its existence!  That’s a little circular.  Instead, we’re going to prove something slightly more general.

Theorem: A real $n\times n$ matrix $A$ is invertible if and only if $A^{m}$ is, for $m\in {\mathbb N}$.

Okay, now, we’re going to prove this two different ways.  One way is sort of the “naive” way, just following your nose — the other is a more sophisticated idea using rank-nullity and determinants.

Proof (Naive!). One direction is trivial; namely, if $A$ is invertible, then $A^{m} = AA\cdots A$ is also invertible, with inverse $A^{-1}A^{-1}\cdots A^{-1} = (A^{m})^{-1}$.

To prove the other direction, let’s suppose $A^{m}$ is invertible.  Then $A^{m}(A^{m})^{-1} = 1$.  This implies $AA^{m-1}(A^{m})^{-1} = 1$.  Grouping these elements like $A(A^{m-1}(A^{m})^{-1}) = 1$ we notice that $A^{m-1}(A^{m})^{-1}$ is the inverse for $A$.  Hence, $A$ is invertible.  $\Diamond$.

This one was nice and elementary, but we can be a bit fancier.  Let’s note three things before we begin the more sophisticated proof.

First, determinants are multiplicative: $det(AB) = det(A)\cdot det(B)$.  If this is not obvious to you, try multiplying two matrices together and take determinants.  There are some nice proofs of this in a number of high school texts as well as a more advanced proof in Dummit and Foote.

Second, we have the chains

$ker(A) \subseteq ker(A^{2}) \subseteq \cdots$

$Im(A) \supseteq Im(A^{2}) \supseteq \cdots$

and these are relatively easy to derive.  Note that if $x\in Ker(A)$, then $Ax = 0$ which implies $A(Ax) = A(0) = 0$, and so the kernel chain holds.  By applying rank-nullity, we obtain the second chain.  We’ll only need to use the first chain, though.

Third, we have that $A$ is invertible if and only if $det(A) \neq 0$ if and only if $ker(A) = \{0\}$.  These can be found in any linear algebra book, but are kind of fun to show yourself.

Proof (Determinants only). Note $det(A) \neq 0$ implies $det(A)det(A)\cdots det(A) = det(A^{m}) \neq 0$ which implies $A^{m}$ is invertible.   The implication also goes the other way.  $\Diamond$.

Proof (Kernels only). Note that one way is trivial.  If $A$ is invertible, then $A^{m}$ is by applying $A^{-1}$ a total of $m$ times to $A^{m}$.

Now suppose $A^{m}$ is invertible.  Then $ker(A^{m}) = 0$.  By our chain, this means that $ker(A^{r}) = 0$ for each $r\leq m$, and hence $A^{r}$ is also invertible.  In particular, $A^{1} = A$ is invertible.  $\Diamond$.

As you can see, there’s a number of really neat proofs for this one, and they’re all essentially equivalent.  Notice that the determinant proof was the shortest, though it takes quite a lot to actually define the determinant and prove everything necessary; in other words, the determinant proof took a lot of building up machinery before we were able to prove it.

On the other hand, the kernel proof was somewhat more slick (using part of our elementary proof to prove one direction) and using a chain to prove the other direction.  Note that it automatically proves that each preceding power is invertible “automatically”, while the elementary proof would have taken a bit more to prove this.