Trace

Suppose \(A=[a_{ij}]\in F^{n\times n}\)
Define:
\(tr(A)=\sum_{i=1}^{n}a_{ii}\)
Example:
\( A = \begin{bmatrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 4 & 5 & 6 \end{bmatrix} , tr(A)= 1 + 8 + 6 = 15 \)

Definition

Suppose \(A,B\in F^{n\times n},\alpha,\beta\in F\)

\( tr(\alpha A \pm\beta B)= \alpha tr(A)\pm \beta tr(B) \)
\( tr(A^T) = tr(A) \)
\( tr(A^H)= \overline{tr(A)}= tr(\overline{A}) \)
\( tr(I_n)= n \)
\( tr(AB)\neq tr(A)tr(B) \)

Theorem

Suppose \(A\in F^{m\times n}, B\in F^{n\times m}\)
\( tr(AB) = tr(BA) \)

Proof:

Suppose \( C=AB \in F^{m\times m}, D=BA \in F^{n\times n} \)
\( tr(AB)=tr(C)\)
\( =\sum_{i=1}^{m}c_{ii}= \sum_{i=1}^{m}\sum_{k=1}^{n}a_{ik}b_{ki} \)
\( =\sum_{k=1}^{n}\sum_{i=1}^{m}b_{ki}a_{ik} =\sum_{k=1}^{n}d_{kk}=tr(D)=tr(BA) \)

Example

Prove that there do not exist \(n\times n\) matrices \(A\) and \(B\) such that \(AB-BA=I_n\).

solution

Using proof by contradiction.
Suppose \( (A,B\in F^{n\times n}) \rightarrow (AB-BA=I_n) \)
\( tr(AB-BA)=tr(I_n) \) checking this equation whether to be right.
\( tr(I_n)= n \)
\( tr(AB-BA)= tr(AB)-tr(BA) \)
\(\because tr(AB)=tr(BA)\)
\(\therefore tr(AB)-tr(BA)=0\)
\( \Rightarrow tr(AB-BA)=tr(I_n) \) is contradiction.
Therefore, there do not exist \(n\times n\) matrices \(A\) and \(B\) such that \(AB-BA=I_n\)

If \( tr(A^{T}A)=0 \), then \( A=O \).

A和自己的轉置矩陣相乘為0, 那A為0

Reference

黃子嘉-線性代數