\({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\)by induction
Consider an orthogonal set of nonzero vectors as \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) and \({c_1},{c_2}\) as a nonzero scalar. Prove that \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2}} \right\}\) is also an orthogonal set. This demonstrates that when the vectors in an orthogonal set are normalized, then the set would still be orthogonal.
If the set \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) is orthonormal and \({\bf{x}} = {c_1}{{\bf{v}}_1} + {c_{12}}{{\bf{v}}_2}\), then the vectors \({c_1}{{\bf{v}}_1}\) and \({c_2}{{\bf{v}}_2}\) are orthogonal.
According to the Pythagoras Theorem and properties of the norm as shown below:
\(\begin{array}{c}{\left\| {\bf{x}} \right\|^2} = {\left\| {{c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2}} \right\|^2}\\ = {\left\| {{c_1}{{\bf{v}}_1}} \right\|^2} + {\left\| {{c_2}{{\bf{v}}_2}} \right\|^2}\\ = {\left( {{c_1}\left\| {{{\bf{v}}_1}} \right\|} \right)^2} + {\left( {{c_2}\left\| {{{\bf{v}}_2}} \right\|} \right)^2}\\ = {\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2}\end{array}\)
Therefore, the stated equality is true for \(p = 2\). Assume that the equality is true for \(p = k\), with \(k \ge 2\).
Consider that \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_{k + 1}}} \right\}\) as an orthonormal set and let \({\bf{x}} = {c_1}{{\bf{v}}_1} + \ldots + {c_k}{{\bf{v}}_k} + {c_{k + 1}}{{\bf{v}}_{k + 1}} = {{\bf{u}}_k} + {c_{k + 1}}{{\bf{v}}_{k + 1}}\), with \({{\bf{u}}_k} = {c_1}{{\bf{v}}_1} + \ldots + {c_k}{{\bf{v}}_k}\).
It is observed that \({{\bf{u}}_k}\) and \({c_{k + 1}}{{\bf{v}}_{k + 1}}\) is orthogonal since \({{\bf{v}}_j} \cdot {{\bf{v}}_{k + 1}} = 0\) for \(j = 1, \ldots ,k\). According to the Pythagoras Theorem and the assumption that the claimed equality is true for \(k\) and since \({\left\| {{c_{k + 1}}{{\bf{v}}_{k + 1}}} \right\|^2} = {\left| {{c_{k + 1}}} \right|^2}{\left\| {{{\bf{v}}_{k + 1}}} \right\|^2} = \left| {{c_{k + 1}}} \right|\)
\(\begin{array}{c}{\left\| {\bf{x}} \right\|^2} = {\left\| {{{\bf{u}}_k} + {c_{k + 1}}{{\bf{v}}_{k + 1}}} \right\|^2}\\ = {\left\| {{{\bf{u}}_k}} \right\|^2} + {\left\| {{c_{k + 1}}{{\bf{v}}_{k + 1}}} \right\|^2}\\ = {\left| {{c_1}} \right|^2} + \ldots + {\left| {{c_{k + 1}}} \right|^2}\end{array}\)
Therefore, the equality holds for \(p = k\) indicates that it's true for \(p = k + 1\). According to the principle of induction, equality holds for all integers \(p \ge 2\).
Hence, it is verified that the equality \({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\) by induction.
