ThmDex – An index of mathematical definitions, results, and conjectures.
Pythagorean theorem
Formulation 0
Let $I$ be an D1247: Inner product-normed vector space such that
(i) $\langle \cdot, \cdot \rangle$ is the D34: Inner product in $I$
(ii) $\Vert \cdot \Vert$ is the D504: Inner product norm in $I$
(iii) $x_1, \dots, x_N \in I$ is an D9: Orthogonal collection in $I$
Then \begin{equation} \left\Vert \sum_{n = 1}^N x_n \right\Vert^2 = \sum_{n = 1}^N \Vert x_n \Vert^2 \end{equation}
Proofs
Proof 0
Let $I$ be an D1247: Inner product-normed vector space such that
(i) $\langle \cdot, \cdot \rangle$ is the D34: Inner product in $I$
(ii) $\Vert \cdot \Vert$ is the D504: Inner product norm in $I$
(iii) $x_1, \dots, x_N \in I$ is an D9: Orthogonal collection in $I$
Applying the additivity of inner product in each argument and orthogonality of the collection $x_1, \ldots, x_N$, we have \begin{equation} \begin{split} \left\Vert \sum_{n = 1}^N x_n \right\Vert^2 & = \left( \sqrt{\left\langle \sum_{n = 1}^N x_n, \sum_{n = 1}^N x_n \right\rangle} \right)^2 \\ & = \left\langle \sum_{n = 1}^N x_n, \sum_{n = 1}^N x_n \right\rangle \\ & = \sum_{n = 1}^N \sum_{m = 1}^N \langle x_n, x_m \rangle = \sum_{n = 1}^N \sum_{m = 1}^N \langle x_n, x_m \rangle I_{\{ n \}}(m) = \sum_{n = 1}^N \langle x_n, x_n \rangle = \sum_{n = 1}^N \Vert x_n \Vert^2 \end{split} \end{equation} where the $I_{\{ n \}}$ are each an D41: Indicator function. $\square$