Question

Suppose that $V$ is an inner product space and $L: V \rightarrow V$ is an isometry, so $\|L[\mathbf{v}]\|=\|\mathbf{v}\|$ for all $\mathbf{v} \in V$. Prove that $L$ also preserves the inner product: $\langle L[\mathbf{v}], L[\mathbf{w}]\rangle=\langle\mathbf{v}, \mathbf{w}\rangle$. Hint: Look at $\|L[\mathbf{v}+\mathbf{w}]\|^2$.

    Suppose that $V$ is an inner product space and $L: V \rightarrow V$ is an isometry, so $\|L[\mathbf{v}]\|=\|\mathbf{v}\|$ for all $\mathbf{v} \in V$. Prove that $L$ also preserves the inner product: $\langle L[\mathbf{v}], L[\mathbf{w}]\rangle=\langle\mathbf{v}, \mathbf{w}\rangle$. Hint: Look at $\|L[\mathbf{v}+\mathbf{w}]\|^2$.

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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 20

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Specifically, for any vectors $\mathbf{v}, \mathbf{w} \in V$, we have: \[ \|L[\mathbf{v} + \mathbf{w}]\| = \|\mathbf{v} + \mathbf{w}\|. \] Show more…

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Suppose that $V$ is an inner product space and $L: V \rightarrow V$ is an isometry, so $\|L[\mathbf{v}]\|=\|\mathbf{v}\|$ for all $\mathbf{v} \in V$. Prove that $L$ also preserves the inner product: $\langle L[\mathbf{v}], L[\mathbf{w}]\rangle=\langle\mathbf{v}, \mathbf{w}\rangle$. Hint: Look at $\|L[\mathbf{v}+\mathbf{w}]\|^2$.

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