Similarity of matrices in an equivalence relation.
Created almost 9 years ago, updated 5 months ago
ID23636
Theorem: Similarity of matrices is an equivalence relation on the set of $n \times n$ matrices.
Proof:
Reflexive: $A = I^{-1}AI$, so $A \sim A$.
Symmetric: If $B = P^{-1}AP$, then $A = PBP^{-1} = (P^{-1})^{-1}B(P^{-1})$, so $A \sim B$.
Transitive: If $B = P^{-1}AP$ and $C = Q^{-1}BQ$, then $C = Q^{-1}P^{-1}APQ = (PQ)^{-1}A(PQ)$, so $A \sim C$.
Since similarity is an equivalence relation, it partitions the set of $n \times n$ matrices into similarity classes. All matrices in the same class share the same:
Eigenvalues
Determinant and trace
Characteristic and minimal polynomials
Jordan canonical form
The Jordan form serves as a canonical representative for each similarity class (over $\mathbb{C}$).
**Theorem:** Similarity of matrices is an **equivalence relation** on the set of $n \times n$ matrices.
**Proof:**
1. **Reflexive:** $A = I^{-1}AI$, so $A \sim A$.
2. **Symmetric:** If $B = P^{-1}AP$, then $A = PBP^{-1} = (P^{-1})^{-1}B(P^{-1})$, so $A \sim B$.
3. **Transitive:** If $B = P^{-1}AP$ and $C = Q^{-1}BQ$, then $C = Q^{-1}P^{-1}APQ = (PQ)^{-1}A(PQ)$, so $A \sim C$.
Since similarity is an equivalence relation, it partitions the set of $n \times n$ matrices into **similarity classes**. All matrices in the same class share the same:
- Eigenvalues
- Determinant and trace
- Characteristic and minimal polynomials
- Jordan canonical form
The Jordan form serves as a canonical representative for each similarity class (over $\mathbb{C}$).