# Properties regarding the trace of a matrix by hedongchenchen

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```									      General Mathematics Vol. 14, No. 4 (2006), 11–14

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Properties regarding the trace of a matrix
Amelia Bucur

s
In memoriam of Associate Professor Ph. D. Luciana Lupa¸

Abstract

There exist in many collection of mathematics problems applica-
tions concerning the trace of a matrix (ex. ...). We understand by
the trace of a matrix the sum of all elements that are on the matrix
regarding the trace of a matrix.

2000 Mathematics Subject Classiﬁcation: 15A60

Through out the paper if A is an n × n matrix, we write tr A to denote
the trace of A and det A for the determinant of A. If A is positive deﬁnite
we write A > 0.
Application 1. Let A ∈ M2 (C) and n ∈ N∗ with An = I2 . Show that if
A + det A is a real matrix, then tr A and det A are real numbers.
Proof. Let f (x) = X n − tr A · X + det A be the characteristic polyoma
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Accepted for publication (in revised form) ....., 2006

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12                                                                                                        Amelia Bucur

of the A matrix and λ1 , λ2 its roots. Because λn = λn = 1, results that
1      2
|λ1 | = |λ2 | = 1. By the hypothesis trA + detA = λ1 + λ2 + λ1 λ2 is a real
1 1
number, so λ1 + λ2 + λ1 λ2 is a real number. We obtain λ1 + λ2 + λ11λ2 from
R, so tr A+1 is real number. Then 1 + tr A+1 = tr A+det A+1 is real number
det A                            det A        det A
and tr A + det A + 1 also, so det A and then tr A is real number.
Application 2. If A > 0 and B > 0, then
0 < tr (AB)m ≤ (tr (AB))m       for all m ∈ N∗
Proof. The equality takes place for n = 1. If n > 1, for B = I the
inequality is true because 0 < tr (An ) ≤ (tr A)n , become
n                     n                m

λm
i      ≤                λi            ,
i=1                    i=1

where λ1 λ2 , . . . , λn are the eigenvalues of A.

If A → AB, the result has been proved.
Application 3. If Ai > 0 and Bi > 0 (i = 1, 2, . . . , k) then
k                 n                    k                              k
tr         Ai Bi           ≤       tr           An
i                tr             Bin
i=1                                i=1                                i=1

If Ai Bi > 0 (i = 1, 2, . . . , k), then
k                 n                    k                              k
tr         Ai Bi           ≤       tr           An
i                tr             Bin
i=1                                i=1                                i=1

Proof. Because
k                          n                              k
n
0 ≤ tr              αAi + Bi                  = α tr                          An
i     +
i=1                                                    i=1

k                                               k
+2αtr            tr         Ai Bi           + tr             tr             Bin
i=1                                          i=1

the result has been proved.
Properties regarding the trace of a matrix                                                 13

In order to demonstrate the second inequality it is suﬃcient to demon-
strate that
k             n            k
(1)                      tr            Ai Bi       ≤   tr         Ai Bi
i=1                        i=1

k
Because Ai Bi > 0 for all i = 1, 2, . . . , k, we have U =                      Ai Bi > 0. The
i=1
n
inequality (1) will result by the fact that tr (U )n ≤ (tr U ) , for all U > 0.
Application 4. If A > 0 and B > 0 then
m
n(det A det B) n ≤ tr (An B n )

for any positive integer m.
Proof. Since A is diagonaligable, there exists an orthodiagonal matrix P
and a diagonal matrix D such that D = P ⊤ A (see [2]). So if the eigenvalues
of A are λ1 , λ2 , . . . , λn , thenD = diag (λ1 , λ2 , . . . , λn ).

Let b11 (m), b22 (m), . . . , bnn (m) denote the elements of (P BP ⊤ )n . Then

1               1                    1
tr (An B n ) = tr (P Dn P ⊤ B n ) = tr (Dn P ⊤ B n P ) =
n               n                    n
1                      1
= tr [Dn (P ⊤ BP )n ] = [λm b11 (m) + λm b22 (m) + . . . + λm bnn (m)].
n
n                      n 1             2

Using the arithmetic - mean geometric - mean inequality, we get
1                                  1                                 1
(2)          tr (An B n ) ≤ [λm λm . . . λm ] n [b11 (m) b22 (m) . . . bnn (m)] n .
1 2         n
n
Since det A ≤ a11 a22 . . . ann for any positive deﬁnite matrix A, we con-
clude that
det (P ⊤ BP )n ≤ b11 (m)b22 (m) . . . bnn (m)

and
det Dn ≤ λm λm . . . λm .
1 2         n
14                                                                Amelia Bucur

Therefore from (2) if fellows that
1                            1                   1
tr (An B n ) ≤ [det (Dn )] n [det (P ⊤ BP )m ] n =
n                                m                 m                 m
= [det (P ⊤ AP )] n [det (P ⊤ BP )] n = (det A det B) n .

Here we used the fact that A > 0 and B > 0. The proof is complete.

Corollary 1.Let A and X be positive deﬁnite n × n - matrices such that
det X = 1. Then
1
n(det A) ≤ tr (AX) .
n
Proof. Take B = X and m = 1 in Applications 4.

References
a a                                                    as
[1] C. N˘st˘sescu, Culegere de probleme de algebr˘, Ed. Didactic˘ ¸i Peda-
a
a        s
gogic˘, Bucure¸ti, 1984 (in Romanian).

a a              ¸a
[2] C. N˘st˘sescu, C. Nit˘, Bazele algebrei, vol. I, Ed. Academiei R.S.R.,
s
Bucure¸ti, 1986 (in Romanian).

”Lucian Blaga” University of Sibiu
Faculty of Sciences
Department of Mathematics
¸
Str. Dr. I. Ratiu, no. 5–7
550012 Sibiu - Romania