Matrix Similarity is Equivalence Relation
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 criteria 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:
\(\ds \mathbf P \mathbf B \mathbf P^{-1}\) | \(=\) | \(\ds \mathbf P \mathbf P^{-1} \mathbf A \mathbf P \mathbf P^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf{I_n} \mathbf A \mathbf{I_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf A\) |
So matrix similarity is symmetric.
$\Box$
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:
\(\ds \mathbf C\) | \(=\) | \(\ds \mathbf P_2^{-1} \mathbf P_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\mathbf P_1 \mathbf P_2}^{-1} \mathbf A \paren {\mathbf P_1 \mathbf P_2}\) | 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.
$\Box$
So, by definition, matrix similarity is an equivalence relation.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices