If we have two $n \times n$ matrices $M$ and $N$, does this mean that I can "wrap" $N$ around $M$, i.e., $M = N^{-1} M N$? If so, what property allows us to do this?

I'm more confident saying that $M = M I_n = M N N^{-1}$, but I don't see what property would allow me to shuffle the $N^{-1}$ to the other side.

  • $\begingroup$ It all depends on $M$. If you know big theorems like Jordan Canonical form then you can calculate the set of all $N$ that commute with $M$ in this way. But here's an easy case: if $M$ has $n$ distinct eigenvalues, then every eigenvector of $M$ is an eigenvector of $N$ $\endgroup$ – ancient mathematician Sep 28 '20 at 10:15

There are two conditions you need here:

  1. Form $N^{-1}MN$ requires matrix $N$ to be invertible.

  2. Multiplying both sides to the $N$ on the left leads to $$NM = NN^{-1}MN = MN,$$ so $M$ and $N$ should commute.


The property you are searching for is called commuting matrices. Two matrices commute if

$$\mathbf{A}\mathbf{B} = \mathbf{B}\mathbf{A}$$

See this question for more details: When is matrix multiplication commutative?. As well as this wiki: https://en.wikipedia.org/wiki/Commuting_matrices

  • $\begingroup$ Right, but in general only two diagonal matrices are commutable. Presumably that means that I can't assume $M = N^{-1}M N$ if at least one of the two is not diagonal? $\endgroup$ – Migwell Sep 28 '20 at 10:18
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    $\begingroup$ @Migwell that is incorrect. Generally, if both matrices have the same eigenvectors, they are commutable. $\endgroup$ – kvantour Sep 28 '20 at 10:19
  • $\begingroup$ Thank you, this actually answers my core misunderstanding about the definition of commutability. $\endgroup$ – Migwell Sep 28 '20 at 10:24
  • $\begingroup$ @Migwell consider matrices $\begin{pmatrix}1 & 2 \\ -2 & 0\end{pmatrix}$ and $\begin{pmatrix}-3 & 2 \\ -2 & -4\end{pmatrix}$ - they are not diagonal but commute and satisfy your condition $\endgroup$ – Anton Grudkin Sep 28 '20 at 10:26

Given that $N$ is invertible, we have the following sequence of equivalent statements:

$$M = N^{-1}MN$$ $$\iff N(M) = N(N^{-1}MN)$$ $$\iff NM = (NN^{-1})MN$$ $$\iff NM = IMN$$ $$\iff NM = MN$$

So if $N$ is invertible, then $M = N^{-1}MN$ if and only if $M$ and $N$ commute.


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