Every $k\times k$ positive matrix $A=(a_{ij})$ can be realised as a Gram matrix, i.e., vectors $x_j$, $1\leq j\leq k$, can be found so that $a_{ij}=\sef{x_i,x_j}$ for all $i,j$.
Solution. By Exercise I.2.2, $A=B^*B$ for some $B$. Let $$\bex B=(x_1,\cdots,x_k). \eex$$ Then $$\bex A=\sex{\sef{x_i,x_j}}. \eex$$
[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.10
原文：http://www.cnblogs.com/zhangzujin/p/4115285.html