Exercise 1.17

Question

Exercise 17: Prove that
    $$|x + y|^2  + |x - y|^2 = 2 |x|^2  + 2|y|^2 $$
    if $x \in R^k$ and $y \in R^k$. Interpret this geometrically, as a statement about parallelograms.

ghostofgarborg · Accepted Answer

(	extit{Matt ``frito'' Lundy}) \
    This solution will use some linear algebra.  In particular, for any inner product $\langle \cdot {,} \cdot angle$ the norm is 
    $$| \cdot | = \sqrt{\langle \cdot {,} \cdot angle}$$
    or
    $$| \cdot |^2 = \langle \cdot {,} \cdot angle.$$
    In this real vector space $R^k$, for any $x,y,z \in R^k$ we also have the following properties:
    $$\langle x + y , z angle = \langle x , z angle + \langle y , zangle $$
    $$\langle x , y angle = \langle y , x angle.$$
    Then
    \begin{align*}
        |x + y|^2  + |x - y|^2 &= \langle x , x angle + 2\langle x , y angle + \langle y , y angle
            + \langle x , x angle - 2\langle x , y angle +\langle y , y angle \
        &= 2\langle x , x angle + 2\langle y , y angle \
        &= 2|x|^2 + 2|y|^2
    \end{align*}
    Geometrically, this says that for any parallelogram in Euclidian Space, the sum of the lengths of the diagonals is equal to the perimeter.

Amit Awasthi · Answer

\begin{proof}
      Consider each term of the left side of the equation:
      \begin{align*}
        |\mathbf{x} + \mathbf{y}|^2 &= (\mathbf{x}+\mathbf{y})\cdot(\mathbf{x}+\mathbf{y})\
        &= \mathbf{x}\cdot\mathbf{x} + 2\mathbf{x}\cdot\mathbf{y} + \mathbf{y}\cdot\mathbf{y},\
        |\mathbf{x} - \mathbf{y}|^2 &= (\mathbf{x}-\mathbf{y})\cdot(\mathbf{x}-\mathbf{y})\
        &= \mathbf{x}\cdot\mathbf{x} - 2\mathbf{x}\cdot\mathbf{y} + \mathbf{y}\cdot\mathbf{y}.
      \end{align*}
      Adding the two together we get
      \begin{equation*}
        |\mathbf{x} + \mathbf{y}|^2 + |\mathbf{x} - \mathbf{y}|^2 = 2|\mathbf{x}|^2 + 2|\mathbf{y}|^2.\qedhere
      \end{equation*}
    \end{proof}
    \begin{remark}
      Geometrically, if we let $\mathbf{x}$ and $\mathbf{y}$ represent the sides of a parallelogram, $\mathbf{x}+\mathbf{y}$ and $\mathbf{x}-\mathbf{y}$ represent the diagonals of the parallelogram. The the area found by summing the squares of the diagonals and the area found by multiplying the sum of the squares of the sides by 2 are the same. When $\mathbf{x}\cdot\mathbf{y} = 0$, i.e., the parallelogram is a rectangle, we get the Pythagorean theorem for two different triangles. When $|\mathbf{x}| = |\mathbf{y}|$ as well, i.e., the parallelogram is a square, we get the Pythagorean theorem multiplied by 2.
    \end{remark}

Exercise 1.17

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Exercise 1.17

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