I am working on an old exam containing a question about Diagonalizable matrix, I am quite confident about the subject overall but there is one simple thing that bothers me, a lot!

We are given the formula $A=PDP^{-1}$ I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it. I have watched a few examples on Wikipedia and in some PDF file that hinted me in the correct direction.

What I believe, For example:

$A=PD$ can be rewritten as $AP^{-1}=D$ if this was regular variables, but they are not, they are matrices, and with matrices, you can only put the "latest" number in front of the equation like this:

$A=PD$ can be rewritten as $P^{-1}A=P^{-1}PD=D$ this works fine but with the formula $A=PDP^{-1}$ I get stuck in an infinite loop of moving things in front of the equation and everything becomes a mess. There is clearly something easy I have missed out. Here are my calculations:




Now I want to get rid of $P^{-1}$ from the right-hand side but since I can only but things in front of D, everything gets messy and I get stuck in a never-ending loop of making things more complicated and adding $PP^{-1}=I$ to the equation hoping it would help but it doesn't :(

  • 1
    $\begingroup$ Right-multiply the equation $P^{-1}A=DP^{-1}$ by $P$,$$P^{-1}AP=DP^{-1}P=D$$Since matrix products don't commute in general, you can either pre-multiply or left-multiply a matrix $X$ with another matrix $Y$, as in $XY$, or you can do post-multiplication or right-multiplication, as in $YX$. $\endgroup$ – Shubham Johri Jan 10 at 10:02
  • $\begingroup$ So you actually are looking for $P$ which will transform $A$ into a diagonal form, right? $\endgroup$ – denklo Jan 10 at 10:07
  • $\begingroup$ @denklo Looking for D, but exact values are not needed, I know how to do that, I do not know how to rewrite matrix equations $\endgroup$ – J. Doe Jan 10 at 10:10
  • $\begingroup$ @ShubhamJohri Okay so the trick is that I can multiply in front as I have done, or at the end of the equation? $\endgroup$ – J. Doe Jan 10 at 10:11
  • 1
    $\begingroup$ That's correct, @J.Doe. $\endgroup$ – Shubham Johri Jan 10 at 10:13

You just have to take into account that matrix multiplication is not commutative.

So from $A=PDP^{-1}$, just multiply with $P^{-1}$ on the left, and with $P$ on the right. As matrix multiplication is associative, you obtain $$P^{-1}AP=P^{-1}(PDP^{-1})P=(P^{-1}P)D(P^{-1}P)=D.$$


I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it.

We start with the identity $A=PDP^{-1}$.

Begin by multiplying both sides on the left by $P^{-1}$ and then on the right by $P$, then make use of the properties that

  • For any invertible matrix $A$, $AA^{-1} = A^{-1}A = I$.
  • For all square matrices $A$, we have $AI=IA=A$.
  • Matrix multiplication is associative, i.e. $ABC = (AB)C = A(BC)$, where the multiplicative is assumed to be defined for the matrices $A,B,C$.

So, we obtain,

$$\begin{align} A=PDP^{-1} &\iff P^{-1}AP=P^{-1}PDP^{-1}P \\ &\iff P^{-1}AP=(P^{-1}P)D(P^{-1}P) = IDI = D \end{align}$$


$$P^{-1}AP = D$$

  • $\begingroup$ Nice thing about this result is that it's not exclusive to matrices at all. As invertible matrices form a Abelian Group over addition and Group over multiplication we see this results holds for all structures that have those properties! $\endgroup$ – DavidG Jan 12 at 9:19

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