VOL. 36, 1950 MA THEMA TICS: F. D. MURNAGHAN 673 SCHWARZ' INEQUALITY AND LORENTZ SPA CES BY FRANcIs D. MURNAGHAN INSTITUTO TECNOL6GICO DE AERONAUTICA, RiO DB JANEIRO Communicated September 1, 1950 The most fundamental formula in the theory of Euclidean or unitary metric spaces is, doubtless, Schwarz' inequality which states that the modulus of the scalar product of two vectors, real or complex, is dominated by the product of their magnitudes (the modulus of the scalar product being equal to the product of the magnitudes when, and only when, the two vectors are linearly dependent). Expressed in matrix notation (x*y)(y*x) % (x*x)(y*y) where x and y are any two n X 1 matrices, real or complex, and x*, y* are the conjugate transposed matrices of x and y, respectively. The proof, now classical, of this fundamental result runs as follows. If X is any complex number the squared magnitude of Xx + y is never negative; on varying the argument of X it follows that neither of the two quadratic polynomial functions of I XI: (X*X) I X 1 2 2 x*y I XI + (y*y) is ever negative. Hence the quadratic polynomial function (x*x)e2 + 21 x*y e + (y*y) of the real variable e is never negative and this implies, on using the theory of the zeros of a quadratic polynomial, Schwarz' inequality. We wish to point out in the present note that the positiveness of x*x and the theory of the zeros of a quadratic polynomial are really somewhat foreign to Schwarz' fundamental inequality and to give the analog of Schwarz' inequality for Lorentz spaces, in which -the scalar product of a non-zero n X 1 matrix by itself may be negative or zero, instead of being always positive as in Euclidean or unitary spaces. Let xi and x2 be any two linearly independent n X 1 matrices, real or complex, and denote by X the n X 2 matrix whose column matrices are xi and x2. If xi and x2 are linear combinations of two other n X 1 matrices, yi and y2 (necessarily linearly independent), we have X = YA where Y is the n X 2 matrix whose column matrices are y, and y2 and A is a non- singular 2 X 2 matrix. We take y, to be a multiple of xl, so chosen that Yi*y = 1 and we set y2 = - (yl*x2)yl + x2. Then Y1*Y2 = 0 and y2 is not the zero matrix. The 2 X 2 matrix Y* Y, being the matrix (O °*Y), has a positive determinant and, since X*X = A *( Y* Y)A, the determinant of the 2 X 2 matrix X*X, being the product of the determinant of the 2 X 2 matrix Y* Y by det A 12, is positive. Since X*X is the matrix (X *X1 Xl*X2 this proves Schwarz' inequality in the case where the X1 X2 X2 two n X 1 matrices xi and x2 are linearly independent. When they are linearly dependent X may be written in the form YA where the 2 X 2 Downloaded by guest on September 30, 2021 674 MA THEMA TICS: F. D. M URNA GHA N PROC. N. A. S. matrix A is singular. Hence the 2 X 2 matrix X*X is singular; this completes the proof of Schwarz' inequality. Before passing to the case of Lorentz spaces we make the obvious remark that if we wish to avail ourselves of the diagonalization property of Her- mitian matrices the above proof of Schwarz' inequality may be given in the following (less elementary and only slightly more concise) variant form. The 2 X 2 matrix X*X is Hermitian and so there exists a 2 X 2 unitary matrix U such that U*(X*X) U is diagonal. Since X*X is posi- tively definite the diagonal elements of this diagonal form of X*X (i.e., the characteristic numbers of X*X) are positive and hence the determinant of X*X is positive. This proves the part of Schwarz' inequality which is concerned with linearly independent n X 1 matrices xl and X2. The sim- plest and most direct proof of the part which is concerned with linearly dependent n X 1 matrices xi and X2 is that given above. Turning now to the case of Lorentz spaces, let F be the diagonal n X n matrix all of whose diagonal elements are 1 save the last which is -1. The Lorentz scalar product of two n X 1 matrices x and y is y*Fx and we say that the n X 1 matrix x, real or complex, is time-like, null or space-like according as x*Fx is negative, zero or positive, respectively. The last element of a time-like n X 1 matrix cannot be zero and it follows that every pair of linearly independent n X 1 matrices xl and X2 contains a space-like linear combination; this implies that if xj*Fxi .Z 0 and xl*Fx2 = O then x2*Fx2 > 0. Let us denote by X the n X 2 matrix whose column matrices are any given pair of linearly independent matrices xl and x2; if xi and x2 are linear combinations of two other n X 1 matrices Yi and y2, necessarily linearly independent, we have X = YA where Y is the n X2 matrix whose column matrices are yi and Y2 and A is a non-singular 2 X 2 matrix. Hence the determinant of the 2 X 2 matrix X*FX has the same sign as the determinant of the matrix Y*FY and, if xl and x2 are linearly dependent, the matrix X*FX is singular. Taking xi and X2 to be linearly independent there exists a linear combination Yi of xl and X2 which is such that yj*Fyi = 1 (since there exists a space-like linear combination of xi and x2). We may suppose that y, is not a multiple of X2 (since it is not a multiple of both xi and X2) and we set Y2 = - (Yl*FX2)Yl + X2. Then YI*FY2 = 0 and the determinant of the 2 X 2 matrix Y*FY is y2*Fy2. Thus, if there exists no time-like linear combination of xl and x2, the determinant of the 2 X 2 matrix X*FX is non-negative. Similarly, if there does exist a time-like linear combination of xl and X2, this determi- nant is negative. Thus we have the following analog, for Lorentz spaces, of Schwarz' inequality: (xj*FxI) (x2*Fx2) < Ixl*Fx21 2 Downloaded by guest on September 30, 2021 VOL. 36, 1950 MA THEMA TICS: F. D. MURNAGHAN 675 if, and only if, there exists a time-like linear combination of the two n X 1 matrices xi and X2, supposed linearly independent. Since there always exists a space-like linear combination of two linearly independent n X 1 matrices, xl and X2, one of the two characteristic num- bers of the 2 X 2 Hermitian matrix X*FX is positive. The second characteristic number may be either positive, zero or negative; in the first case, i.e., when the 2 X 2 Hermitian matrix X*FX is positively def- inite, we say that the n X 2 matrix X is space-like (so that every non- trivial linear combination of the column matrices of a space-like n X 2 matrix is space-like). Thus we may complete the analog, for Lorentz spaces, of Schwarz' inequality as follows: If (xi*Fxl)(x2*Fx2) = xl*Fx212 there exists a null non-trivial linear combination of xi and X2 while there does not exist a time-like linear combi- nation of xi and x2; (xi*Fxl)(x2*Fx2) > IX1*Fx2 12 if, and only if, the n X 2 matrix X, whose column matrices xl and X2 are linearly independent, is space-like. These results may be extended to the case of p n X 1 matrices xi, x. where 3 < p < n. The proof given above shows that the p X P matrix X*FX, where X is the n X p matrix whose column matrices are xI, ... xp, is singular if the p n X 1 matrices xl, .. ., x. are linearly dependent and so we shall suppose that they are linearly independent. The p X p matrix X*FX cannot have less than p - 1 positive characteristic numbers (since every pair of linearly independent n X 1 matrices possesses a space-like linear combination). Thus the n X p matrix X is space-like (by which we mean that every non-trivial linear combination of its colunm matrices is space-like) if, and only if, the determinant of the p X p matrix X*FX is positive. This determinant is negative if, and only if, there exists a time- like linear combination of the p n X 1 matrices xi, . ., x.. The matrix X*FX is singular if, and only if, there exists a null non-trivial linear combi- nation of xi, . ., x, (while no time-like linear combination of these exists). In the case p = n the determinant of X*FX is negative since there certainly exists a time-like linear combination of n linearly independent n X 1 matrices xi, . ., xn; the negativeness of det X*FX is a priori evident since det X*FX is the product of det F by det X 2. If we shall ever have to consider physical spaces with more than one time direction we must introduce quasi-Lorentz spaces.