The Pythagorean Theorem in N-Space

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

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

(5.1)

This relationship generalizes to

dimensions, as we can easily show:

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

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

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 , , in which case

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

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 or 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.