Free Books    Mathematics of the DFT  

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

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

This relationship generalizes to $ N$ 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 $ 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

$\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 $ 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:

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

Next Section:
Projection
Previous Section:
Orthogonality