44 Matrix trace

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March 14, 2024

The trace of a square $n \times n$ matrix $A$ is defined as the sum of the diagonal components of the matrix, i.e., $Tr (A) = \sum_{i = 1}^{n} A_{i i} .$

44.1 Properties of trace

For any $A \in R^{n \times m}$ and $B \in R^{m \times n}$ , $Tr (A B) = Tr (B A) .$

An immediate consequence of the above is that for any $A \in R^{n \times m}$ , $B \in R^{m \times ℓ}$ , and $C \in R^{ℓ \times n}$ , $Tr (A B C) = Tr (B C A) = Tr (C A B) .$
For $A, B \in R^{n \times n}$ and any scalars $α$ and $β$ , $Tr (α A + β B) = α Tr (A) + β Tr (B) .$
In fact, it can be shown that trace is the only function $f$ that satisfies the following three properties: for any $A, B \in R^{n \times n}$ and scalars $α$ , $β$ :
- $f (α A + β B) = α f (A) + β f (B)$ .
- $f (A B) = f (B A)$ .
- $f (I) = n$ .
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., $0 \leq Tr (A B) \leq Tr (A) Tr (B) .$

Thus, $Tr (A^{2}) \leq (Tr (A))^{2}$ .
The above can be generalized to arbitrary powers. For positive semidefinite matrices (of the same size), and for any positive integer $m$ : $0 \leq Tr ((A B)^{m}) \leq (Tr (A^{2 m}))^{1 / 2} (Tr (B^{2 m}))^{1 / 2}$ and $Tr ((A B)^{m}) = (Tr (A B))^{m} .$
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 $\frac{1}{p_{1}} + \dots + \frac{1}{p_{m}} = 1.$ Let $| A | : = (A^{⊺} A)^{1 / 2}$ Then, $Tr (| A_{1} A_{2} \dots A_{m} |) \leq \prod_{i = 1}^{m} (Tr (A_{i}^{p_{i}}))^{1 / p_{i}}$ and $Tr (| A_{1} A_{2} \dots A_{m} |) \leq Tr (\sum_{i = 1}^{m} \frac{1}{p_{i}} A_{i}^{p_{i}}) .$
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 $n \times n$ matrices and $p_{1}, \dots, p_{m}$ are positive numbers such that $\frac{1}{p_{1}} + \dots + \frac{1}{p_{m}} = 1.$ Let $| A | : = (A^{⊺} A)^{1 / 2}$ Then, for any integer $r \geq 1$ , $\begin{matrix} (H \ddot{o} lder inequality) & Tr ({| A_{1} A_{2} \dots A_{m} |}^{r}) \leq \prod_{i = 1}^{m} (Tr ({| A_{i} |}^{r p_{i}}))^{1 / p_{i}} \end{matrix}$ and $\begin{matrix} (Young inequality) & Tr ({| A_{1} A_{2} \dots A_{m} |}^{r}) \leq Tr (\sum_{i = 1}^{m} \frac{1}{p_{i}} {| A_{i} |}^{r p_{i}}) . \end{matrix}$

Notes

See Yang and Feng (2002) and Shebrawai and Albadawani (2012) for generalizations of the above inequalities.