How do we algebraically manipulate non-commutative equations? (Ie. Diagonalization: $A=PDP^{-1} \rightarrow AP=PD$)

Question

1. How can we manipulate non-commutative equations? (See question 3 below)

2. It seems in the formula for Diagonalization, $A=PDP^{-1} \rightarrow AP=PD$, does one just multiply both sides by P?

3. Main question: How do we manipulate it instead to $PAP^{-1}=D$?

Answer 1

First of all, it's not the equation that is non-commutative, but rather the operation used in the equation.

As to the main point, there's actually nothing special here, we just have to be a bit more careful. The basic logical apparatus of equality (see here for more on this) tells us that given any well-defined function $F$ and any true equation $s=t$, the equation $F(s)=F(t)$ is again a true equation. In particular, this applies to $F(x)=xy$ for a fixed $y$, as well as to $G(x)=yx$ for a fixed $y$. The non-commutativity of multiplication in our current setting means that $F$ and $G$ are not the same in general, but that's fine.

So, looking at your particular example, we argue as follows:

• We start with $A=PDP^{-1}$ by assumption.

• We now apply the principle above with $F(x)=xP$. This gives $$F(A)=F(PDP^{-1})\quad\implies \quad AP=PDP^{-1}P\quad\implies AP=PD.$$

Now we are still being a bit informal - we're being rather cavalier with respect to the associative property of multiplication (which still holds). Ultimately a fully formal approach will involve some parentheses-juggling. But the above is perfectly adequate for the topic.

Note that we didn't simply "multiply by $P$" - in the absence of commutativity, "multiplication by $P$" isn't well-defined. Our $F$ in this case was multiplication on the right by $P$. We could also have multiplied by $P$ on the left on both sides of the equation, getting $$PA=P^2DP^{-1}.$$ However, we could not multiply by $P$ on the left in the left hand side and on the right on the right hand side: $PA=PD$ is not something we can conclude from $A=PDP^{-1}$ (without further hypotheses, at least).