Question

Suppose $J, M: U \rightarrow U$ are self-adjoint linear functions on an inner product space $U$. (a) Prove that $\langle J[\mathbf{u}], \mathbf{u}\rangle=\langle M[\mathbf{u}], \mathbf{u}\rangle$ for all $\mathbf{u} \in U$ if and only if $J=M$. (b) Explain why this result is false if the self-adjointness hypothesis is dropped.

    Suppose $J, M: U \rightarrow U$ are self-adjoint linear functions on an inner product space $U$.
(a) Prove that $\langle J[\mathbf{u}], \mathbf{u}\rangle=\langle M[\mathbf{u}], \mathbf{u}\rangle$ for all $\mathbf{u} \in U$ if and only if $J=M$.
(b) Explain why this result is false if the self-adjointness hypothesis is dropped.

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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 19

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Step 1

Since $J$ and $M$ are self-adjoint, $N$ is also self-adjoint. This follows because the adjoint of a difference of two operators is the difference of their adjoints, i.e., $N^* = J^* - M^* = J - M = N$. Show more…

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Suppose $J, M: U \rightarrow U$ are self-adjoint linear functions on an inner product space $U$. (a) Prove that $\langle J[\mathbf{u}], \mathbf{u}\rangle=\langle M[\mathbf{u}], \mathbf{u}\rangle$ for all $\mathbf{u} \in U$ if and only if $J=M$. (b) Explain why this result is false if the self-adjointness hypothesis is dropped.

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