34. Definite Hermitian forms

A Hermitian form is positive definite, if for all values of the variable always ³ 0 and = 0 only for = 0. Definite negative forms are defined correspondingly. At times, Hermitian forms H are referred to as being positive and negative, respectively.

If is positive definite, certainly the determinant |H| 0, since otherwise the homogeneous, linear system of equations H = 0 would have a solution
= 0 and therefore , although a positive definite form can only vanish for = 0. The value of a determinant of a negative definite form must differ from zero for the same reason. Since the determinant |H| equals the product of the characteristic roots of H, all characteristic roots of a [positive or negative] definite Hermitian matrix are not zero.

By means of the principal axes transformation, every Hermitian form can be given the form (32.1) in which the characteristic roots g₁, ··· , g_n appear as coefficients. However, a form like (32.1) is positive and negative definite if and only if all the characteristic roots g_n are positive and negative, respectively.

The Hermitian form is positve (negative) if and only if all roots of H are positive (negative).

For a determination of whether a given form is definite, this criterion is, in general, not suitable, because it is laborious to find out whether an algebraic equation of degree n (that is, the characteristic equation of H) has only positive or negative roots.

In order to derive a more readily handled criterion, you can proceed as follows: If is positive definite, as we have just found out, all characteristic roots of H are positive, whence the determinant |H| as product of the characteristic roots is also positive. More over, consider the sectional determinant

of the matrix H. It shows that for a positive definite form all sectional determinants are positive. In fact, setting x_k+1 = ··· = x_n = 0, you obtain from a Hermitian form in the variables x₁, ··· , x_k , which obviously is also positive definite. A matrix of this form arises from H by omission of the last n-k rows and columns, whence its determinant is H_k and positive like the determinant of every positive definite form.

The condition for a Hermitian form to be positive definite when all its sectional determinants are positive is formulated in

Theorem 20

The Hermitian form is positive definite if and only if all sectional determinants H₁, ··· , H_n are positive; it is negative definite if the signs of the sequence H₁, ··· , H_n alternate and H₁ < 0.

Proof: The first part of the rule relating to positive definite forms will be proved first. It has already been shown that in the case of a positive definite form all secitonal determinants are positive. The converse must still be confirmed.

Start from the Hermitean matrix

,

which differs from H only by the bottom right element. As you readily find by expansion with respect to the last column,

moreover, the Rank of B is n - 1, because the sectional determinant H_n-1 is positive.

Now proceed in a similar manner as during the proof of Theorem 19, where by a transformation matrix, one column of which was an eigenvector, one step was taken towards the diagonal form. The difference between the two proofs is that, in general, the transformation is now not unitary and the eigenvector occurs in the last column.

First of all, you must find an eigenvector. Since |B| = 0, B has the number 0 as a characteristic root. The eigenvector corresponding to 0 is a solution of the system B = 0. Due to the Rank B = n - 1 and H_n-1 0, the last of the first n-1 row of B depends linearly on the first n - 1 rows, so that you can omit the last equations from the system of equations B = 0. Hence is determined by the system of equations

h₁₁z₁ + ··· + h_1nz_n = 0,
······································ ···········
h_{n -1,1}z₁ + ··· + h_{n -1,n}z_n = 0.

In view of an observation in Section 27, z₁, ··· ,z_n are proportional to the sequence H_n1, ··· ,H_nn of the adjoint minor determinant of the last row of B or, what is the same, of H. Since H_nn = H_n-1 0, you can use

z₁ = H_n1 /H_n-1, z₂ = H_n2 /H_n-1, ··· , z_n-1 = H_n,n-1 /H_n-1, z_n = 1.

Introduce now into the Hermitian form the new varibles

x₁ = y_i + H_ni y/H_n-1 (i = 1, , n - 1),
x_n = y_n.

The Matrix T of this substitution has as last column the eigenvector , whence While evaluating , note that due to the equation B= 0 there occur only zeroes in the last column of BT, whence also the last column of has only zeroes. Because is a Hermitian matrix, there must also occur only zeroes in its last row, whence

Hence also

If now y_n = 0, then x_i = y_i (i = 1, ··· , n - 1) and

Thus a_ik = h_ik (i,k = 1, ··· , n - 1) and (34.1) yields

On the right hand side of this equation is a Hermitian form in n - 1 variables, the matrix of which arises by stretching of the last row and column. You can split off this form by a corresponding procedure H_n-1|y_n-1|²/H_n-2 and obtain a Hermitian form in n - 2 variables, the matrix of which arises from H by omission of the last two rows and columns. Continuing in this manner, you arrive eventually at

You have again obtained a representation of the Hermitian form as a sum of purely quadratic terms; however, it is, in general, not an unitary, but only some non-singular transformation of the variables. This representation yields now the first half of Theorem 20. In fact, if all the H₁, ··· ,H_n are positive, then obviously the right hand side of (34.2) is 0. The equality sign can only arise when w₁ = ··· = w_n = 0; however, since the w_i are linked to the x_i by a non-singular, linear transformation, also all the x_i must vanish.

The second part of Theorem 20 can be reduced to the just concluded case of a positive definite form. In fact, if you multiply all coefficients of a negative definite form by -1, there arises a positive definite form. In the process, the sectional determinant H_k is just multiplied by (-1)^k.

Exercises

25. Execute the principal axes transformation for the Hermitian matrix

26. Find the largest and smallest value of the real quadratic form , when the variables are subject to the supplementary condition .

Answers

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