Linear Algebra Done Right · Problem 5D.9

Inner Product Spaces — Sheldon Axler

Mathematics — Linear Algebra · diagonalizable-operators

Problem

Suppose R,T∈L(F3)R, T \in \mathcal{L}(\mathbf{F}^3) each have 2,6,72, 6, 7 as eigenvalues. Prove that there exists an invertible operator S∈L(F3)S \in \mathcal{L}(\mathbf{F}^3) such that R=S−1TSR = S^{-1}TS.
Intuitive explanation

Two operators with the same eigenvalues (counting multiplicity) are similar—they represent the same linear transformation in different coordinate systems. The key insight is that both RR and TT can be diagonalized to the same diagonal matrix DD with entries 2,6,72, 6, 7; if R=P−1DPR = P^{-1}DP and T=Q−1DQT = Q^{-1}DQ, then composing these relationships gives you the conjugating operator SS that transforms one into the other.

Intuitive explanation

Two operators with the same eigenvalues (counting multiplicity) are similar—they represent the same linear transformation in different coordinate systems. The key insight is that both RR and TT can be diagonalized to the same diagonal matrix DD with entries 2,6,72, 6, 7; if R=P−1DPR = P^{-1}DP and T=Q−1DQT = Q^{-1}DQ, then composing these relationships gives you the conjugating operator SS that transforms one into the other.

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