41 Matrix trace
The trace of a square \(n × n\) matrix \(A\) is defined as the sum of the diagonal components of the matrix, i.e., \[ \TR(A) = \sum_{i=1}^n A_{ii}. \]
41.1 Properties of trace
For any \(A \in \reals^{n × m}\) and \(B \in \reals^{m × n}\), \[ \TR(AB) = \TR(BA). \]
An immediate consequence of the above is that for any \(A \in \reals^{n × m}\), \(B \in \reals^{m × \ell}\), and \(C \in \reals^{\ell × n}\), \[ \TR(ABC) = \TR(BCA) = \TR(CAB). \]
For \(A, B \in \reals^{n × n}\) and any scalars \(α\) and \(β\), \[ \TR(αA + βB) = α \TR(A) + β \TR(B). \]
In fact, it can be shown that trace is the only function \(f\) that satisfies the following three properties: for any \(A,B \in \reals^{n × n}\) and scalars \(α\), \(β\):
- \(f(αA + βB) = αf(A) + βf(B)\).
- \(f(AB) = f(BA)\).
- \(f(I) = n\).
For a square matrix, trace is equal to the sum of the eigenvalues, i.e., \[ \TR(A) = \sum_{i = 1}^n λ_i(A). \]
For positive semidefinite matrices (of the same size), trace is sub-multiplicative, i.e., \[ 0 \le \TR(AB) \le \TR(A)\TR(B). \]
Thus, \(\TR(A^2) \le (\TR(A))^2\).
The above can be generalized to arbitrary powers. For positive semidefinite matrices (of the same size), and for any positive integer \(m\): \[ 0 \le \TR((AB)^m) \le (\TR(A^{2m}))^{1/2} (\TR(B^{2m}))^{1/2} \] and \[ \TR((AB)^m) = (\TR(AB))^m. \]
Another generalization is the following. Let \(A_1,\dots,A_m\) be positive semidefinite matrices (of the same size) and \(p_1,\dots, p_m\) are positive numbers such that \[ \frac{1}{p_1} + \cdots + \frac{1}{p_m} = 1. \] Let \(\ABS{A} \coloneqq (A^\TRANS A)^{1/2}\) Then, \[ \TR(\ABS{A_1A_2 \cdots A_m}) \le \prod_{i=1}^m (\TR(A^{p_i}_i))^{1/p_i} \] and \[ \TR(\ABS{A_1A_2 \cdots A_m}) \le \TR\biggl(\sum_{i=1}^m \frac{1}{p_i} A_i^{p_i}\biggr). \]
The next result removes the restriction to positive semidefinite matrices (at the cost of working with absolute values). Let \(A_1,\dots,A_m\) be arbitrary \(n × n\) matrices and \(p_1,\dots, p_m\) are positive numbers such that \[ \frac{1}{p_1} + \cdots + \frac{1}{p_m} = 1. \] Let \(\ABS{A} \coloneqq (A^\TRANS A)^{1/2}\) Then, for any integer \(r \ge 1\), \[\begin{equation} \TR(\ABS{A_1A_2 \cdots A_m}^r) \le \prod_{i=1}^m (\TR(\ABS{A_i}^{r p_i}))^{1/p_i} \tag{H\"older inequality} \end{equation}\] and \[\begin{equation} \TR(\ABS{A_1A_2 \cdots A_m}^r) \le \TR\biggl(\sum_{i=1}^m \frac{1}{p_i} \ABS{A_i}^{rp_i}\biggr). \tag{Young inequality} \end{equation}\]
Notes
See Yang and Feng (2002) and Shebrawai and Albadawani (2012) for generalizations of the above inequalities.