91Ó°ÊÓ

To prove that "is similar to" is an equivalence relation on \(\mathrm{M}_{n \times n}(F)\), we need to show that it satisfies reflexivity, symmetry, and transitivity. Two matrices A and B are similar if there exists an invertible matrix P such that \(B=P^{-1}AP\). 1. For reflexivity, consider the identity matrix \(I_n\) and notice that \(A = I_n^{-1}AI_n\), showing that every matrix A is similar to itself. 2. For symmetry, if A is similar to B, then \(B=P^{-1}AP\), with P invertible. Since \(P^{-1}\) is also invertible, we can set \(Q=P^{-1}\) and find that \(A = Q^{-1}BQ\), showing that B is similar to A. 3. For transitivity, assume A is similar to B and B is similar to C, with invertible matrices \(P_1\) and \(P_2\) such that \(B=P_{1}^{-1}AP_{1}\) and \(C=P_{2}^{-1}BP_{2}\). Set \(R=P_{2}P_{1}\), then \(C = R^{-1}AR\), showing that A is similar to C. Since reflexivity, symmetry, and transitivity are satisfied, "is similar to" is an equivalence relation on \(\mathrm{M}_{n \times n}(F)\).