Let \(V\) be a real or complex vector space with an inner product. Show that the quadratic form determined by the inner product satisfies the parallelogram law \[ ||\alpha + \beta||^2 + ||\alpha - \beta||^2 = 2||\alpha||^2 + 2||\beta||^2. \]
Solution
By definition \(\|\gamma\|^2=(\gamma\mid\gamma)\). Therefore \[ \begin{aligned} \|\alpha+\beta\|^2 &=(\alpha\mid\alpha)+(\alpha\mid\beta) +(\beta\mid\alpha)+(\beta\mid\beta),\\ \|\alpha-\beta\|^2 &=(\alpha\mid\alpha)-(\alpha\mid\beta) -(\beta\mid\alpha)+(\beta\mid\beta). \end{aligned} \] Adding cancels the two mixed terms and yields \[ \boxed{ \|\alpha+\beta\|^2+\|\alpha-\beta\|^2 =2\|\alpha\|^2+2\|\beta\|^2.} \] The computation uses only additivity and conjugate symmetry, so it is valid for both real and complex inner-product spaces [@HK71].