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(a) Let $L_1: U \rightarrow V_1$ and $L_2: U \rightarrow V_2$ be linear maps between inner product spaces, with $V_1, V_2$ not necessarily the same. Let $J_1=L_1^* \circ L_1, J_2=L_2^* \circ L_2$. Show that the sum $J=J_1+J_2$ can be written as a self-adjoint combination $J=L^* \circ L$ for some linear operator

    (a) Let $L_1: U \rightarrow V_1$ and $L_2: U \rightarrow V_2$ be linear maps between inner product spaces, with $V_1, V_2$ not necessarily the same. Let $J_1=L_1^* \circ L_1, J_2=L_2^* \circ L_2$. Show that the sum $J=J_1+J_2$ can be written as a self-adjoint combination $J=L^* \circ L$ for some linear operator
 
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Applied Linear Algebra (Undergraduate Texts in Mathematics)
Applied Linear Algebra (Undergraduate Texts in Mathematics)
Peter J. Olver,… 2nd Edition
Chapter 7, Problem 23 ↓

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Recall that for a linear map \( L: U \rightarrow V \), the adjoint \( L^* \) is defined as the unique linear map \( L^*: V \rightarrow U \) such that \( \langle L(u), v \rangle_V = \langle u, L^*(v) \rangle_U \) for all \( u \in U \) and \( v \in V \). Given \(  Show more…

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(a) Let $L_1: U \rightarrow V_1$ and $L_2: U \rightarrow V_2$ be linear maps between inner product spaces, with $V_1, V_2$ not necessarily the same. Let $J_1=L_1^* \circ L_1, J_2=L_2^* \circ L_2$. Show that the sum $J=J_1+J_2$ can be written as a self-adjoint combination $J=L^* \circ L$ for some linear operator
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Key Concepts

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Factorization of Positive Semi-Definite Operators
Positive semi-definite operators are those for which the quadratic form ?x, Ax? is non-negative for all x, and they can always be factored as L*L for some linear operator L. This factorization is valuable because it provides structural insights and simplifications, ensuring that combinations of such operators (when also self-adjoint) can be expressed in the form L*L. This result is fundamental to understanding the spectral properties and stability of various operator equations.
Direct Sum of Inner Product Spaces and Operators
The concept of the direct sum of inner product spaces allows one to combine different spaces into one larger space where the inner product is defined component-wise. This construction is useful in the analysis and synthesis of operators, as it can be used to create combined operators from individual ones. In the context of the problem, the operator L is constructed by considering the direct sum of two linear maps, linking them into a single framework that retains the properties of self-adjointness and positive semi-definiteness.
Adjoint Operator
The adjoint of a linear operator L between inner product spaces is an operator L* that satisfies the property ?Lx, y? = ?x, L*y? for all x and y in the appropriate spaces. This concept is fundamental because it provides a way to relate the action of L on one space to an operator acting on the original space, and it is essential in exploring symmetries and self-adjointness in operators.
Linear Maps Between Inner Product Spaces
Linear maps in the context of inner product spaces are functions that preserve the vector space structure while possibly relating different spaces. Their analysis often involves their behavior with respect to the inner product, leading to considerations of orthogonality, projections, and other geometric properties. Using inner products also allows one to define adjoint operators, which are crucial for many constructions in linear algebra and functional analysis.
Self-Adjoint Operator
An operator is self-adjoint if it equals its own adjoint; that is, A = A*. Self-adjoint operators are particularly important because they have real eigenvalues and exhibit properties similar to symmetric matrices, making them central in spectral theory and applications that involve energy considerations and stability analysis. They also possess a rich theory regarding their diagonalization and factorization.

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Let (V,⟨·,·⟩) be an inner product space, and let β = {v1,...,vn} be an orthonormal basis for an inner product space V. Let T : V → V be a linear operator. Define the map U : V → V by U(x) = ∑_{j=1}^{n} ⟨x, T(vj)⟩ vj, x ∈ V. 1. Prove that U is a linear operator. 2. Prove that U = T*.

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