Matrix Similarity is Equivalence Relation

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Matrix similarity is an equivalence relation.

Proof 1

Follows directly from Matrix Equivalence is Equivalence Relation.


Proof 2

Checking in turn each of the critera for equivalence:


$\mathbf A = \mathbf{I_n}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all order $n$ square matrices $\mathbf A$.

So matrix similarity is reflexive.



Let $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.

As $\mathbf P$ is invertible, we have:

\(\displaystyle \mathbf P \mathbf B \mathbf P^{-1}\) \(=\) \(\displaystyle \mathbf P \mathbf P^{-1} \mathbf A \mathbf P \mathbf P^{-1}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \mathbf{I_n} \mathbf A \mathbf{I_n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \mathbf A\) $\quad$ $\quad$

So matrix similarity is symmetric.



Let $\mathbf B = \mathbf P_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf P_2^{-1} \mathbf B \mathbf P_2$.

Then $\mathbf C = \mathbf P_2^{-1} \mathbf P_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$.

The result follows from the definition of invertible matrix, that the product of two invertible matrices is itself invertible.

So matrix similarity is transitive.


So, by definition, matrix similarity is an equivalence relation.