Exercise 5.1.4

Question

Prove that for vectors in an inner product space
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\begin{equation*}
    ||\mathbf{x \pm y}||^2 = ||\mathbf{x}||^2 + ||\mathbf{y}||^2 \pm 2 	ext{Re}(\mathbf{x,y}).
\end{equation*}
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Recall that Re$(z) = \frac{1}{2}(z + \overline{z})$.

Jing Huang · Accepted Answer

\begin{proof}
\begin{equation*}
\begin{split}
    ||\mathbf{x - y}||^2 &= (\mathbf{x-y}, \mathbf{x-y})  \
    &= (\mathbf{x, x-y}) - (\mathbf{y, x-y})    \
    &= (\mathbf{x,x}) - (\mathbf{x,y}) - (\mathbf{y,x}) + (\mathbf{y,y})  \
    &= ||\mathbf{x}||^2 + ||\mathbf{y}||^2 - (\mathbf{x,y}) - \overline{(\mathbf{x,y})}  \
    &= ||\mathbf{x}||^2 + ||\mathbf{y}||^2 - 2 	ext{Re}(\mathbf{x,y}).  \
\end{split}
\end{equation*}
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Similarly $||\mathbf{x + y}||^2 = ||\mathbf{x}||^2 + ||\mathbf{y}||^2 + 2 	ext{Re}(\mathbf{x,y})$.
\end{proof}

Exercise 5.1.4

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Exercise 5.1.4

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