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Exercise 7.C.11

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Proof. Let $e_{1}, e_{2}, e_{3}$ and $f_{1}, f_{2}, f_{3}$ be orthonormal bases of $𝔽^{3}$ consisting of eigenvectors of $T_{1}$ and $T_{2}$ , respectively, corresponding to the eigenvalues $2, 5, 7$ . Define $S$ by

S e_{j} = f_{j}

for $j = 1, 2, 3$ . One easily checks that $S$ is an isometry (using the Pythagorean Theorem). Then, because $S^{- 1} = S^{∗}$ (by 7.42), we have $S^{∗} f_{j} = e_{j}$ . Thus

T_{1} e_{1} = 2 e_{1} = S^{∗} (2 f_{1}) = S^{∗} (T_{2} f_{1}) = S^{∗} T_{2} S e_{1} .

Similarly $T_{1} e_{2} = S^{∗} T_{2} S e_{2}$ and $T_{1} e_{3} = S^{∗} T_{2} S e_{3}$ . Therefore $T_{1} = S^{∗} T_{2} S$ . □

celiopassos

2017-10-06 00:00

Exercise 7.C.11

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