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Exercise 7.C.1

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Proof. We give a counterexample. Define $T \in L (R^{2})$ by

\begin{aligned} T e_{1} = e_{1} \\ T e_{2} = - e_{2} \end{aligned}

where $e_{1}, e_{2}$ is the standard basis of $ℝ^{2}$ . The matrix of $T$ with respect to this same basis is

(\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}),

which equals its transpose, therefore $T$ is self-adjoint. Moreover, the basis $\frac{1}{\sqrt{2}} (e_{1} + e_{2}), \frac{1}{\sqrt{2}} (e_{1} - e_{2})$ is orthonormal and

\begin{aligned} ⟨ T (\frac{1}{\sqrt{2}} (e_{1} + e_{2})), \frac{1}{\sqrt{2}} (e_{1} + e_{2}) ⟩ & = 0 \\ ⟨ T (\frac{1}{\sqrt{2}} (e_{1} - e_{2})), \frac{1}{\sqrt{2}} (e_{1} - e_{2}) ⟩ & = 0, \end{aligned}

but $T$ is not positive because

⟨ T e_{2}, e_{2} ⟩ = ⟨ - e_{2}, e_{2} ⟩ = - 1 .

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celiopassos

2017-10-06 00:00

Exercise 7.C.1

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