44  Matrix trace

Updated

March 14, 2024

The trace of a square n×n matrix A is defined as the sum of the diagonal components of the matrix, i.e., Tr(A)=i=1nAii.

44.1 Properties of trace

  1. For any ARn×m and BRm×n, Tr(AB)=Tr(BA).

    An immediate consequence of the above is that for any ARn×m, BRm×, and CR×n, Tr(ABC)=Tr(BCA)=Tr(CAB).

  2. For A,BRn×n and any scalars α and β, Tr(αA+βB)=αTr(A)+βTr(B).

  3. In fact, it can be shown that trace is the only function f that satisfies the following three properties: for any A,BRn×n and scalars α, β:

    • f(αA+βB)=αf(A)+βf(B).
    • f(AB)=f(BA).
    • f(I)=n.
  4. For a square matrix, trace is equal to the sum of the eigenvalues, i.e., Tr(A)=i=1nλi(A).

  5. For positive semidefinite matrices (of the same size), trace is sub-multiplicative, i.e., 0Tr(AB)Tr(A)Tr(B).

    Thus, Tr(A2)(Tr(A))2.

  6. The above can be generalized to arbitrary powers. For positive semidefinite matrices (of the same size), and for any positive integer m: 0Tr((AB)m)(Tr(A2m))1/2(Tr(B2m))1/2 and Tr((AB)m)=(Tr(AB))m.

  7. Another generalization is the following. Let A1,,Am be positive semidefinite matrices (of the same size) and p1,,pm are positive numbers such that 1p1++1pm=1. Let |(A|:=(AA)1/2 Then, Tr(|(A1A2Am|)i=1m(Tr(Aipi))1/pi and Tr(|(A1A2Am|)Tr(i=1m1piAipi).

  8. The next result removes the restriction to positive semidefinite matrices (at the cost of working with absolute values). Let A1,,Am be arbitrary n×n matrices and p1,,pm are positive numbers such that 1p1++1pm=1. Let |(A|:=(AA)1/2 Then, for any integer r1, (Ho¨lder inequality)Tr(|(A1A2Am|r)i=1m(Tr(|(Ai|rpi))1/pi and (Young inequality)Tr(|(A1A2Am|r)Tr(i=1m1pi|(Ai|rpi).

Notes

See Yang and Feng () and Shebrawai and Albadawani () for generalizations of the above inequalities.

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