The short answer is no. Assuming a complex n-dimensional Hilbert space, this can be directly appreciated by realising the fact that a Gram matrix G in this space can be written as the product of a matrix with its Hermitian conjugate: G = A† A. Considering an arbitrary n-vector v in this space, one has:
v†. G . v = (A v) †. (A v) ≥ 0. (QED)
To appreciate the equality G = A† A, take n linearly independent vectors in the Hilbert space under consideration and, after indexing them with the indices 1,2, ..., n, construct the corresponding Gram matrix; the element (G)i,j is equal to the inner product of the i-th vector with the j-th vector from the latter ordered set. Now take an orthonormal basis and represent the above-mentioned linearly-independent vectors with respect to the latter basis. You will immediately realise that the i-th column of the matrix A consists of the components of the i-th vector in the adopted ordered set of n linearly independent vectors with respect to the latter orthonormal basis set. Incidentally, this approach shows that the matrix A in G = A† A is unique only up to a unitary matrix. This is not surprising, since with U denoting an n x n unitary matrix, satisfying U† U = I, where I denotes the n x n unit matrix, one can equally express G as G = B† B, where B = U A.
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