On the Proof of Bushell-Trustrum Inequality

Foundation item: Support by the Natural Science Foundation of Henan (0611052600), the Basic and the Forward Position Technology Research Project of Henan. Abstract Bushell and Trustrum (Bushell, 1990, p. 173-178) give the famous Bushell-Trustrum inequality, but their proof exists two main mistakes which make their proof process can not establish. This paper corrects these mistakes and gives the correct proof.

In 1990, Bushell and Trustrum proved Whereas the result proved by Lieb and Thirring, Bushell and Trustum only need to prove ) in their proof firstly, then tr(AB 1 ) k , tr(AB 2 ) k are the smallest and largest values of tr(AB) k , here U i is unitary matrix.The mistakes in their proof are mainly in the following two points: (1) Exist unitary matrix X with rank n, such that X * AX, X * B 1 X, X * B 2 X become diagonal at the same time; (2) ¢ www.ccsenet.org/jmrISSN: 1916-9795 We will point out that unitary matrix X with rank n, which makes X * AX, X * B 1 X, X * B 2 X become diagonal at the same time does not definitely exist, and for general positive semi-definite Hermite matrix, (1) does also not definitely exist.
We give the following conclusions: Exist unitary matrixX with rank n, such that X * AX and X * B 1 X become diagonal at the same time; Exist unitary matrixY with rank n, such that Y * AY and Y * B 2 Y become diagonal at the same time.And for B 1 , B 2 , Thus complete the certification of Bushell-Trustrum inequality.We need to use the following Lemma:

Our proof
Suppose A > 0, B > 0, otherwise for any c > 0, There must be A + cI > 0, B + cI > 0, finally we take limit to the result obtained when c → 0, then we conclude the proof.
Since entire unitary matrix with rank n constitutes a closed set and mapping U → tr(AU BU * ) k is a continuous function defined on this closed set, so there must be the smallest and largest values in U 1 and U 2 , Then Especially, in (4), take U = I, then we have ), we will prove first: Exist unitary matrix X with rank n, such that X * AX and where R, F are n × n rectangular matrix, 0, I are zero matrix and unit matrix on some degree.
Obviously, R is an unitary matrix, and to infinitely small ε 0, R can denoted as Here o(|ε| 2 ) is n × n rectangular matrix, everyone of its element is infinitesimal of higher order of |ε|.For convenient we use o(|ε| 2 ) to denote either matrix or number.

March, 2009
In fact, From mathematical analysis, when x → 0, so elements in ( 11) and ( 12) are infinitesimal of higher order of |ε|, thus (10) holds.For any unitary matrix T, define Since R is unitary matrix, T RT * is unitary matrix.Because B is positive semi-definite Hermite matrix, T R * T * = (T RT * ) * , B is positive semi-definite Hermite matrix too.From (10), we get Here It is easy to prove that for any two unitary matrix with rank n P and Q, have Then from ( 16), ( 18) Here We can prove that (AB) k−1 A ≥ 0 In fact, notice that A and B are both positive semi-definite Hermite matrixes.
It can be proved by induction.In(20), because (AB) k−1 A is positive semi-definite, T is any unitary matrix, so we can choose unitary matrix T, such that D becomes diagonal, The last equation is right because C 1 is Hermite matrix, and By ( 22) and ( 19), then This formula is correct on arbitrary semi-positive Hermite matrix B and infinitely small ε 0. Similarly, we take R, F such that their i, j(i < j) row and column have form of (8), ( 9), and similar to the proof above, then it can be obtained.
Set ε = ηc i j , η > 0, Use the same method c i j = c ji = 0 can be obtained. Suppose Here C i is positive semi-definite Hermite matrix with rank n Set X = T V, then X is an unitary matrix, and This is a diagonal matrix, its diagonal elements are eigenvalues of B 1 , furthermore The last equation is correct because V and D are block matrix with same degree.By ( 28) and (29) we know (30) then by definition of B 1 and (2), we obtain εc 12 + εc 12 = η|c 12 | 2 = 0, then c 12 = c 21 = 0.