About: http://webisa.webdatacommons.org/219841830

Property	Value
http://www.w3.org/1999/02/22-rdf-syntax-ns#type	http://www.w3.org/ns/prov#Entity
http://www.w3.org/ns/prov#value	As far as I can tell, your proof assumes that A is a diagonal matrix so that the eigenvectors are [1, 0, ..., 0], [0, 1, 0, ..., 0], ..., [0, ..., 0, 1] and thus $x^TAx = \sum_{i=1}^n \lambda_i X_i = \lambda_k X_k = \lambda_k$ is true.
http://www.w3.org/ns/prov#wasQuotedFrom	stackexchange.com