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@jemoka / Jemoka Knowledge Base / wiki/concepts/orthonormal_basis.md
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--- title: "orthonormal basis" type: concept related: [Orthonormal, Orthonormal Basis, Norm, Basis, Orthogonal] source: https://www.jemoka.com/posts/kbhorthonormal_basis/ confidence: high status: active --- An Orthonormal basis is defined as a basis of a finite-dimensional vector space that’s orthonormal. Additional Information orthonormal list of the right length is a basis An orthonormal list is linearly independent, and linearly independent list of length dim V are a basis of V. \(\blacksquare\) Writing a vector as a linear combination of orthonormal basis According to Axler, this result is why there’s so much hoopla about orthonormal basis. Result and Motivation For any basis of \(V\), and a vector \(v \in V\), we by basis spanning have: \begin{equation} v = a_1e_1 + \dots a_{n}e_{n} \end{equation} Yet, for orthonormal basis, we can actually very easily know what the \(a_{j}\) are (and not just that some \(a_{j}\) exist). Specifically: \begin{equation} a_{j} = \langle v,e_{j} \rangle \end{equation} That is, for orthonormal basis \(e_{j}\) of \(V\), we have that: \begin{equation} v = \langle v, e_{1} \rangle e_{1} + \dots + \langle v, e_{n} \rangle e_{n} \end{equation} for all \(v \in V\). Furthermore: \begin{equation} \|v\|^{2} = | \langle v,e_1 \rangle|^{2} + \dots + | \langle v, e_{n} \rangle|^{2} \end{equation} Proof Given \(e_{j}\) are basis (nevermind orthonormal quite yet), we have that: \begin{equation} v = a_1e_{1} + \dots + a_{n}e_{n} \end{equation} WLOG let’s take \(\langle v, e_{j} \rangle\): \begin{equation} \langle v,e_{j} \rangle = \langle a_1e_1 + \dots +a_{n}e_{n}, e_{j} \rangle \end{equation} Given additivity and homogenity in the first slot, we now have: \begin{equation} \langle v, e_{j} \rangle = a_{1}\langle e_1, e_{j} \rangle + \dots +a_{n}\langle e_{n}, e_{j} \rangle \end{equation} Of course, each \(e_{i}\) and \(e_{j}\) are orthogonal, so for the most part \(a_{i}\langle e_{i}, e_{j} \rangle = 0\) for \(i \neq j\). Except where \(a_{j} \langle e_{j}, e_{j} \rangle = a_{j} 1 = a_{j}\) because the \(e\) vectors are also norm 1. Therefore: \begin{equation} \langle v, e_{j} \rangle= 0 + \dots +a_{j} + \dots +0 = a_{j} \end{equation} We now have \(\langle v,e_{j} \rangle = a_{j}\) WLOG for all \(j\), as desired. Plugging this in for each \(a_{j}\) and applying Norm of an Orthogonal Linear Combination yields the \(\|v\|^{2}\) equation above. \(\blacksquare\)