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Exercise 7.C.12

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Proof. Let $e_{1}, e_{2}, e_{3}, e_{4}$ denote an orthonormal basis of $𝔽^{4}$ . Define $T_{1}, T_{2} \in L (𝔽^{4})$ by

\begin{aligned} T_{1} e_{1} & = 2 e_{1} \\ T_{1} e_{2} & = 2 e_{2} \\ T_{1} e_{3} & = 5 e_{3} \\ T_{1} e_{4} & = 7 e_{4} \\ T_{2} e_{1} & = 2 e_{1} \\ T_{2} e_{2} & = 5 e_{2} \\ T_{2} e_{3} & = 5 e_{3} \\ T_{2} e_{4} & = 7 e_{4} . \end{aligned}

Then both $T_{1}$ and $T_{2}$ are self-adjoint (the matrices equal their transposes) and $2, 5, 7$ are their eigenvalues. Suppose by contradiction that $S$ is an isometry on $V$ such that $T_{1} = S^{∗} T_{2} S$ . Let $v \in V$ be the vector that $S$ maps to $e_{2}$ . Then

T_{1} v = S^{∗} T_{2} Sv = S^{∗} T_{2} e_{2} = 5 S^{∗} e_{2} = 5 v

Therefore $v \in E (T_{1}, 5) = span (e_{3})$ . Let also $w \in V$ be the vector that $S$ maps to $e_{3}$ . Note that $v, w$ is linearly independent, because $e_{2}, e_{3}$ is linearly independent. Then

T_{1} w = S^{∗} T_{2} Sw = S^{∗} T_{2} e_{3} = 5 S^{∗} e_{3} = 5 w .

Therefore $w \in E (T_{1}, 5) = span (e_{3})$ . But this is a contradiction, because we can’t have a linearly independent list of length $2$ , $v, w$ , in a $1$ -dimensional vector space, $span (e_{3})$ . Hence, there does not exist such $S$ .

Notice that it wasn’t necessary to require $S$ to be an isometry, we just needed to suppose, by contradiction, the existence of an invertible $S$ such that $T_{1} = S^{- 1} T_{2} S$ . This $S$ does not exist. Since the desired isometry must satisfy the same property (because the adjoint of an isometry equals its inverse), it follows that there cannot exist such isometry. The key idea here is that the eigenspaces of $T_{1}$ and $T_{2}$ don’t fit. □

celiopassos

2017-10-06 00:00

Exercise 7.C.12

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