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I was working on the following problem when I stumbled upon an oddity.

If $X=P^{-1}AP$ and $A^3=I$, prove that $X^3=I$

My first approach was to cube both side which led to the following:

$$X^3=(P^{-1}AP)^3$$ $$X^3=(P^{-1})^3A^3P^3$$ Since $A^3=I$ $$X^3=(P^{-1})^3I^3P^3$$ $$X^3=(P^{-1})^3P^3$$ $$X^3=(P^{-1}P)^3$$ $$X=I^3=I$$ However this method seemed to contradict what I had previously learned. Instead I tried this approach: $$X=P^{-1}AP$$ Now premultiply each side by X and we get: $$X^2=P^{-1}APP^{-1}AP=P^{-1}AIAP=P^{-1}A^2P$$

We do this once more: $$X^3=P^{-1}A^2PP^{-1}AP=P^{-1}A^2IAP=P^{-1}A^3P$$

Since $A^3=I$ $$X^3=P^{-1}IP=P^{-1}P=I$$

Does anyone know which approach is correct?

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The latter is correct, of course. – Sangchul Lee Dec 21 '11 at 3:51
The first one is wrong because it treats exponentiation as distributive over multiplication... which it isn't, unless the multiplication is itself commutative. That is, $(AB)^3 = ABABAB \neq A^3 B^3$ in general. – mjqxxxx Dec 21 '11 at 4:03
up vote 5 down vote accepted

It is false in general that $(ABC)^3 = A^3B^3C^3$. This only holds if the elements commute. So you cannot go from $$(P^{-1}AP)^3$$ to $$(P^{-1})^3 A^3 P^3$$ in general, since the equality will not hold if $A$ does not commute with $P$.

Fortunately, in general we have that $$(P^{-1}AP)^n = P^{-1}A^nP$$ is true for all positive integers $n$ (and even for negative ones, if $A$ is invertible). This can be shown by induction on $n$ for positive integers: it is true for $n=1$, since $P^{-1}AP = P^{-1}AP$. Assuming the result is true for $n$, we have $$\begin{align*} (P^{-1}AP)^{n+1} &= (P^{-1}AP)^n(P^{-1}AP)\\ &= (P^{-1}A^nP)(P^{-1}AP)\\ &= P^{-1}A^n(PP^{-1})AP\\ &= P^{-1}A^nAP\\ &= P^{-1}A^{n+1}P \end{align*}$$

One can also see this by verify that the map $B\mapsto P^{-1}BP$ is actually a semigroup homomorphism on the multiplicative semigroup of matrices: for any $B$ and $C$, we have that $$P^{-1}(BC)P = (P^{-1}BP)(P^{-1}CP),$$ as can easily be verified. In fact, it's a ring homomorphism, since $$P{-1}(B+C)P = (P^{-1}BP) + (P^{-1}CP),$$ so conjugation behaves pretty well. But you cannot do what you did first.

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The second approach is correct.

The identity $(AB)^n=A^nB^n$ holds if $AB=BA$, but not typically otherwise. It coincidentally works out in this case that assuming $(ABC)^3=A^3B^3C^3$ leads to the same answer. In general, $(P^{-1}AP)^n=P^{-1}A^nP$, not $(P^{-1})^nA^nP^n$.

I want to add a further note on the importance of commutativity and distinguishing between $(AB)^n$ versus $A^nB^n$. Namely, if $A$ and $B$ are invertible, then $(AB)^2=A^2B^2$ if and only if $AB=BA$.

In one direction, if $AB=BA$, then $(AB)^2=ABAB=A(BA)B=A(AB)B=A^2B^2$. In the other direction, if $(AB)^2=A^2B^2$, then we have:

$\begin{align*} ABAB&=AABB\\ A^{-1}(ABAB)B^{-1}&=A^{-1}(AABB)B^{-1}\\ IBAI&=IABI\\ BA&=AB. \end{align*}$

It is important that $A$ and $B$ are invertible, not just to make the preceding proof work, but because it is possible for $(AB)^2=A^2B^2$ to hold when $A$ and $B$ do not commute. An example of this is given by $A=\begin{bmatrix}0&1\\0&1\end{bmatrix}$ and $B=\begin{bmatrix}1&1\\0&0\end{bmatrix}$. But such cases are atypical, and in the absence of commutativity, for each $n>1$, $(AB)^n=ABABAB\cdots AB$ and $A^nB^n=AAA\cdots ABBB\cdots B$ cannot be considered equal unless proven so.

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