📂Matrix Algebra

대각합

Definition

Let the matrix $A$ be given as follows.

$$A= \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}$$

The sum of the diagonal elements of $A$ is defined as the trace of $A$, and it is denoted as follows.

$$\text{tr}(A)=\text{Tr}(A)=a_{11}+a_{22}+\cdots + a_{nn}=\sum \limits_{i=1}^{n} a_{ii}$$

Explanation

The trace can also be considered as a function in the following way. Let $M_{n\times n}(\mathbb{R})$ be the set of $n\times n$ matrices with real components. Then, $\text{Tr}$ is a function defined as follows.

$$\text{Tr} : M_{n\times n} (\mathbb{R}) \to \mathbb{R},\quad \text{Tr}(A)=\sum \limits_{i=1}^{n} a_{ii}$$

If the trace is a function, there’s no reason it can’t be differentiated.

Properties

Let $A,B,C$ be a $n \times n$ matrix and $k$ be a constant.

(a) The trace of a scalar multiple is the same as the scalar multiple of the trace.

$$\text{Tr}(kA)= k\text{Tr}(A)$$

(b) The trace of a sum is the same as the sum of the traces.

$$\text{Tr}(A+B)=\text{Tr}(A)+\text{Tr}(B)$$

(a)+(b) The trace is linear.

$$\text{Tr}(kA+B)=k\text{Tr}(A)+\text{Tr}(B)$$

(c) The trace of $AB$ is the same as the trace of $BA$.

$$\text{Tr}(AB) = \text{Tr}(BA)$$

• (c’) Cyclic Property: From the above fact, it can be inferred that the following equation holds.

$$\text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB)$$

• (c") It actually holds for any $A \in \mathbb{R}^{n \times k}$ and $B \in \mathbb{R}^{k \times n}$ as well.

(d) The trace of $A$ is the same as the trace of $A^{T}$.

$$\text{Tr}(A) = \text{Tr}(A^{T})$$

Proof