Exercise 5.1.7

Question

Prove the parallelogram identity for an inner product space $V$,
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\begin{equation*}
    ||\mathbf{x+y}||^2 + ||\mathbf{x-y}||^2 = 2(||\mathbf{x}||^2 + ||\mathbf{y}||^2).
\end{equation*}
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Jing Huang · Accepted Answer

\begin{proof}
\begin{equation*}
\begin{split}
    ||\mathbf{x+y}||^2 + ||\mathbf{x-y}||^2 &= (\mathbf{x+y}, \mathbf{x+y}) + (\mathbf{x-y}, \mathbf{x-y})  \
    &= (\mathbf{x,x}) + (\mathbf{x,y}) + (\mathbf{y,x}) + (\mathbf{y,y}) + \
    &\quad \; (\mathbf{x,x}) - (\mathbf{x,y}) - (\mathbf{y,x}) + (\mathbf{y,y})  \
    &= 2(\mathbf{x,x}) + 2(\mathbf{y,y})  \
    &= 2(||\mathbf{x}||^2 + ||\mathbf{y}||^2).
\end{split}
\end{equation*}
\end{proof}

Exercise 5.1.7

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Exercise 5.1.7

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