Let 3C be hilbert space (any dimensionality, real or complex scalars). Let P be a hermitian projection. Let A be any hermitian operator. The compression of A to P3C [3] is PAP, considered as an operator on P3R. Compressions of completely continuous positive operators are of interest in connection with estimating eigenvalues: the Fischer-Courant nminimax theorem [5, p. 2351 says the kth highest eigenvalue of PAP is not greater than that of A. Compressions enter in the study of more general mappings of operators, often via Nalmark's theorem [6]. Especially in the first connection, the case where PJC is finitedimensional is interesting. But in some problems a finite-dimensional subspace may be known, not via the operator P, but via an arbitrary set of vectors which span it; if they are not orthonormal, one would rather not have to find P. This suggests that the following elementary formulas may be worth pointing out. I suppose that at least Formula 1 must be known already, but not, apparently, very widely. NOTATION. xi, * ,X form a linear basis of P3R. G denotes the determinant of their Gramian (n X n matrix with i, j entry (xi, xj)). If the kth row of the Gramian is replaced by (z, xi), . . . , (z, x.), all other rows being left unchanged, the determinant of the resulting matrix will be denoted G(xk; z). Evident properties: G(xk; z) =0 if z= (1 -P)z or z=xi (i%k), while G(xk; Xk) =G; also G(xk; z) is linear in z. These may be summed up by saying that G(xk; z) = (z, x)G, where {x1, *, xI} is the basis of P3R biorthonormal with { xi, n, x FORMULA 1. Pz=G-1 Gk G(xk; Z)Xk. (This notation here and below means summation over all available values of the index.) PROOF. Uniquely z= Es aix+(1 -P)z. Substitute this on both sides, and use the evident properties of G(xk; z). FORMULA 2. tr(PAP) =G ZEk G(xk; Axk). PROOF. Let 1, * * * , in be orthonormal eigenvectors of PAP, and Xi, * * * 9 Xn their respective eigenvalues; then xi= E, Ti,ip, where T is some nonsingular matrix. Recall that