The Pythagorean Theorem in N-Space

In 2D, the Pythagorean Theorem says that when $\underline{x}$ and $\underline{y}$ are orthogonal, as in Fig. 5.8, (i.e., when the triangle formed by $\underline{x}$, $\underline{y}$, and $\underline{x}+\underline{y}$, with $\underline{y}$ translated to the tip of $\underline{x}$, is a right triangle), then we have

$$\Vert\underline{x}+\underline{y}\Vert^2 = \Vert\underline{x}\Vert^2 + \Vert\underline{y}\Vert^2 . \protect$$ (5.1)

This relationship generalizes to $N$ dimensions, as we can easily show:

$$\Vert\underline{x}+\underline{y}\Vert^2 = \left<\underline{x}+\underline{y},\underline{x}+\underline{y}\right>$$

$$= \left<\underline{x},\underline{x}\right>+\left<\underline{x},\und... ...eft<\underline{y},\underline{x}\right>+\left<\underline{y},\underline{y}\right>$$

$$= \Vert\underline{x}\Vert^2 + \left<\underline{x},\underline{y}\rig... ...\overline{\left<\underline{x},\underline{y}\right>} + \Vert\underline{y}\Vert^2$$

$$= \Vert\underline{x}\Vert^2 + \Vert\underline{y}\Vert^2 + 2 \left\{\left<\underline{x},\underline{y}\right>\right\} \protect$$ (5.2)

If $\underline{x}\perp \underline{y}$, then $\left<\underline{x},\underline{y}\right>=0$ and Eq. (5.1) holds in $N$ dimensions.

Note that the converse is not true in ${\bf C}^N$. That is, $\Vert\underline{x}+\underline{y}\Vert^2 = \Vert\underline{x}\Vert^2 + \Vert\underline{y}\Vert^2$ does not imply $\underline{x}\perp \underline{y}$ in ${\bf C}^N$. For a counterexample, consider $\underline{x}= (j,1)$, $\underline{y}= (1, -j)$, in which case

$$\Vert\underline{x}+\underline{y}\Vert^2 = \Vert 1+j,1-j\Vert^2 = 4 = \Vert\underline{x}\Vert^2 + \Vert\underline{y}\Vert^2$$

while $\left<\underline{x},\underline{y}\right> = j\cdot 1 + 1 \cdot\overline{-j} = 2j$.

For real vectors $\underline{x},\underline{y}\in{\bf R}^N$, the Pythagorean theorem Eq. (5.1) holds if and only if the vectors are orthogonal. To see this, note that, from Eq. (5.2), when the Pythagorean theorem holds, either $\underline{x}$ or $\underline{y}$ is zero, or $\left<\underline{x},\underline{y}\right>$ is zero or purely imaginary, by property 1 of norms (see §5.5.2). If the inner product cannot be imaginary, it must be zero.

Note that we also have an alternate version of the Pythagorean theorem:

$$\underline{x}\perp y \;\implies\; \Vert\underline{x}-y\Vert^2 = \Vert\underline{x}\Vert^2 + \Vert y\Vert^2$$