The Pythagorean Theorem in N-Space

In 2D, the Pythagorean Theorem says that when $x$ and $y$ are orthogonal, as in Fig.5.8, (i.e., when the vectors $x$ and $y$ intersect at a right angle), then we have

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

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

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

$$= \left<x,x\right>+\left<x,y\right>+\left<y,x\right>+\left<y,y\right>$$

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

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

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

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

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

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

For real vectors $x,y\in{\bf R}^N$, 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 $x$ or $y$ is zero, or $\left<x,y\right>$ is zero or purely imaginary, by property 1 of norms (see §5.8.2). If the inner product cannot be imaginary, it must be zero.

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

$$x\perp y\,\,\implies\,\, \Vert x-y\Vert^2 = \Vert x\Vert^2 + \Vert y\Vert^2$$