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 criteria 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:

\(\ds \mathbf P \mathbf B \mathbf P^{-1}\) \(=\) \(\ds \mathbf P \mathbf P^{-1} \mathbf A \mathbf P \mathbf P^{-1}\)
\(\ds \) \(=\) \(\ds \mathbf{I_n} \mathbf A \mathbf{I_n}\)
\(\ds \) \(=\) \(\ds \mathbf A\)

So matrix similarity is symmetric.



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


\(\ds \mathbf C\) \(=\) \(\ds \mathbf P_2^{-1} \mathbf P_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2\)
\(\ds \) \(=\) \(\ds \paren {\mathbf P_1 \mathbf P_2}^{-1} \mathbf A \paren {\mathbf P_1 \mathbf P_2}\) Inverse of Matrix Product

By Product of Matrices is Invertible iff Matrices are Invertible, $\paren {\mathbf P_1 \mathbf P_2}$ is invertible .

So matrix similarity is transitive.


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