Chapter 13: Problem 57
Show that self-adjoint, skew-adjoint, and unitary (orthogonal) operators are normal.
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Chapter 13: Problem 57
Show that self-adjoint, skew-adjoint, and unitary (orthogonal) operators are normal.
These are the key concepts you need to understand to accurately answer the question.
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Prove Theorem 13.1: Let \(T\) be a linear operator on an \(n\) -dimensional inner product space \(V\). Then (a) There exists a unique linear operator \(T^{*}\) on \(V\) such that \\[\langle T(u), v\rangle=\left\langle u, T^{*}(v)\right\rangle \quad \text { for all } u, v \in V\\] (b) Let \(A\) be the matrix that represents \(T\) relative to an orthonormal basis \(S=\left\\{u_{i}\right\\} .\) Then the conjugate transpose \(A^{*}\) of \(A\) represents \(T^{*}\) in the basis \(S\)
Show that the sum of two positive (positive definite) operators is positive (positive definite).
\(\operatorname{Let} A=\left[\begin{array}{ll}2 & i \\ i & 2\end{array}\right] .\) Verify that \(A\) is normal. Find a unitary matrix \(P\) such that \(P^{*} A P\) is diagonal. Find \(P^{*} A P\).
Let \(V\) be an inner product space, and suppose \(U: V \rightarrow V\) (not assumed linear) is surjective (onto) and preserves inner products; that is, \(\langle U(v), U(w)\rangle=\langle u, w\rangle\) for every \(v, w \in V\). Prove that \(U\) is linear and hence unitary.
Prove Theorem 13.2: (i) \(\left(T_{1}+T_{2}\right)^{*}=T_{1}^{*}+T_{2}^{*}\) (iii) \(\left(T_{1} T_{2}\right)^{*}=T_{2}^{*} T_{1}^{*}\) (ii) \((k T)^{*}=\bar{k} T^{*}\), (iv) \(\left(T^{*}\right)^{*}=T\). (i) For any \(u, v \in V\), $$ \begin{aligned} \left\langle\left(T_{1}+T_{2}\right)(u), v\right\rangle &=\left\langle T_{1}(u)+T_{2}(u), \quad v\right\rangle=\left\langle T_{1}(u), v\right\rangle+\left\langle T_{2}(u), v\right\rangle \\ &=\left\langle u, T_{1}^{*}(v)\right\rangle+\left\langle u, T_{2}^{*}(v)\right\rangle=\left\langle u, T_{1}^{*}(v)+T_{2}^{*}(v)\right\rangle \\ &=\left\langle u, \quad\left(T_{1}^{*}+T_{2}^{*}\right)(v)\right\rangle \end{aligned} $$ The uniqueness of the adjoint implies \(\left(T_{1}+T_{2}\right)^{*}=T_{1}^{*}+T_{2}^{*}\). (ii) For any \(u, v \in V\), $$ \langle(k T)(u), \quad v\rangle=\langle k T(u), v\rangle=k\langle T(u), v\rangle=k\left\langle u, T^{*}(v)\right\rangle=\left\langle u, \bar{k} T^{*}(v)\right\rangle=\left\langle u,\left(\bar{k} T^{*}\right)(v)\right\rangle $$ The uniqueness of the adjoint implies \((k T)^{*}=\bar{k} T^{*}\). (iii) For any \(u, v \in V\), $$ \begin{aligned} \left\langle\left(T_{1} T_{2}\right)(u), v\right\rangle &=\left\langle T_{1}\left(T_{2}(u)\right), v\right\rangle=\left\langle T_{2}(u), T_{1}^{*}(v)\right\rangle \\ &=\left\langle u, T_{2}^{*}\left(T_{1}^{*}(v)\right)\right\rangle=\left\langle u,\left(T_{2}^{*} T_{1}^{*}\right)(v)\right\rangle \end{aligned} $$ The uniqueness of the adjoint implies \(\left(T_{1} T_{2}\right)^{*}=T_{2}^{*} T_{1}^{*}\). (iv) For any \(u, v \in V\), $$ \left\langle T^{*}(u), v\right\rangle=\overline{\left\langle v, T^{*}(u)\right\rangle}=\overline{\langle T(v), u\rangle}=\langle u, T(v)\rangle $$ The uniqueness of the adjoint implies \(\left(T^{*}\right)^{*}=T\).
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