Question

A matrix A ? Mn×n(R) is Gramian if there is a B ? Mn×n(R) such that A = BtB. Prove that A is Gramian if and only if A is symmetric and all of its eigenvalues are non-negative. Hint: For (?), note that A is diagonalizable via an orthonormal basis {u1 , . . . , un } where ui is an eigenvector of A with eigenvalue ?i. Consider the linear operator T on Rn where T(ui) = ??iui. Now take B = [T]std and check that A = BtB.