The Pythagorean Theorem for Inner Product Spaces

Recall the Pythagorean Theorem from geometry that says if we have a right triangle (that is, a triangle that contains a right angle), then the sum of the squares of the lengths of the shortest two sides of this triangle is equal to the square of the length of the longest side of the triangle (the hypotenuse), that is if $c$ is the hypotenuse of a triangle, then:

(1)

\begin{equation} a^2 + b^2 = c^2 \end{equation}

What's fascinating is that the Pythagorean theorem can be extended to inner product spaces in terms of norms.

Theorem 1 (The Pythagorean Theorem for Inner Product Spaces): Let $V$ be an inner product space. If $u, v \in V$ are orthogonal to each other then $\| u + v \|^2 = \| u \|^2 + \| v \|^2$.

• Proof: Let $u, v \in V$. Since $u$ is orthogonal to $v$ we have that $<u, v> = 0 = <v, u>$. Thus:

(2)

\begin{align} \quad \| u + v \|^2 = <u + v, u + v> \\ \quad \| u + v \|^2 = <u, u> + <u, v> + <v, u> + <v, v> \\ \quad \| u + v \|^2 = <u, u> + <v, v> \\ \quad \| u + v \|^2 = \| u \|^2 + \| v \|^2 \end{align}

• Thus our proof is complete. $\blacksquare$

In $\mathbb{R}^2$, if the vectors $u$ and $v$ are orthogonal to each other, the Pythagorean theorem can be geometrically interpreted as the sum of the lengths $\| u \|$, $\| v \|$ of the vectors $u$ and $v$ squared is equal to the length of the vector sum $\| u + v \|$ squared.