Matrix Similarity is Equivalence Relation

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Theorem

Matrix similarity is an equivalence relation.


Proof 1

Follows directly from Matrix Equivalence is Equivalence Relation.

$\blacksquare$


Proof 2

Checking in turn each of the critera for equivalence:


Reflexive

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

So matrix similarity is reflexive.

$\Box$


Symmetric

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.

$\Box$


Transitive

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.

$\Box$


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

$\blacksquare$


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