tr AB = tr BA的证明

突然忘记怎么证明 trAB=trBAtr AB= tr BAtrAB=trBA，于是自己推导了一遍。记录下来，强化记忆。

开始证明

首先给定两个nnn 阶方阵
A=(a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮an1an2⋯ann)B=(b11b12⋯b1nb21b22⋯b2n⋮⋮⋱⋮bn1bn2⋯bnn)A=\begin{pmatrix} {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{pmatrix}\quad B=\begin{pmatrix} {b_{11}}&{b_{12}}&{\cdots}&{b_{1n}}\\ {b_{21}}&{b_{22}}&{\cdots}&{b_{2n}}\\ {\vdots}&{\vdots}&{\ddots}&{\vdots}\\ {b_{n1}}&{b_{n2}}&{\cdots}&{b_{nn}}\\ \end{pmatrix} A=⎝⎜⎜⎜⎛a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann⎠⎟⎟⎟⎞B=⎝⎜⎜⎜⎛b11b21⋮bn1b12b22⋮bn2⋯⋯⋱⋯b1nb2n⋮bnn⎠⎟⎟⎟⎞算算 trABtr ABtrAB
AB=(a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮an1an2⋯ann)⋅(b11b12⋯b1nb21b22⋯b2n⋮⋮⋱⋮bn1bn2⋯bnn)AB=\begin{pmatrix} {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{pmatrix}\cdot \begin{pmatrix} {b_{11}}&{b_{12}}&{\cdots}&{b_{1n}}\\ {b_{21}}&{b_{22}}&{\cdots}&{b_{2n}}\\ {\vdots}&{\vdots}&{\ddots}&{\vdots}\\ {b_{n1}}&{b_{n2}}&{\cdots}&{b_{nn}}\\ \end{pmatrix} AB=⎝⎜⎜⎜⎛a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann⎠⎟⎟⎟⎞⋅⎝⎜⎜⎜⎛b11b21⋮bn1b12b22⋮bn2⋯⋯⋱⋯b1nb2n⋮bnn⎠⎟⎟⎟⎞trAB=∑j=1na1jbj1+∑j=1na2jbj2+⋯∑j=1nanjbjn=∑i=1n∑j=1naijbjitr AB=\sum_{j=1}^{n}a_{1j}b_{j1}+\sum_{j=1}^{n}a_{2j}b_{j2}+\cdots\sum_{j=1}^{n}a_{nj}b_{jn}\\ =\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}b_{ji} trAB=j=1∑na1jbj1+j=1∑na2jbj2+⋯j=1∑nanjbjn=i=1∑nj=1∑naijbji再算算 trBAtr BAtrBA
BA=(b11b12⋯b1nb21b22⋯b2n⋮⋮⋱⋮bn1bn2⋯bnn)⋅(a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮an1an2⋯ann)BA=\begin{pmatrix} {b_{11}}&{b_{12}}&{\cdots}&{b_{1n}}\\ {b_{21}}&{b_{22}}&{\cdots}&{b_{2n}}\\ {\vdots}&{\vdots}&{\ddots}&{\vdots}\\ {b_{n1}}&{b_{n2}}&{\cdots}&{b_{nn}}\\ \end{pmatrix}\cdot \begin{pmatrix} {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{pmatrix} BA=⎝⎜⎜⎜⎛b11b21⋮bn1b12b22⋮bn2⋯⋯⋱⋯b1nb2n⋮bnn⎠⎟⎟⎟⎞⋅⎝⎜⎜⎜⎛a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann⎠⎟⎟⎟⎞trBA=∑j=1nb1jaj1+∑j=1nb2jaj2+⋯∑j=1nbnjajn=∑i=1n∑j=1nbijajitr BA=\sum_{j=1}^{n}b_{1j}a_{j1}+\sum_{j=1}^{n}b_{2j}a_{j2}+\cdots\sum_{j=1}^{n}b_{nj}a_{jn}\\ =\sum_{i=1}^{n}\sum_{j=1}^{n}b_{ij}a_{ji} trBA=j=1∑nb1jaj1+j=1∑nb2jaj2+⋯j=1∑nbnjajn=i=1∑nj=1∑nbijaji

问题关键

原命题也就是要证明
∑i=1n∑j=1naijbji=∑i=1n∑j=1nbijaji\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}b_{ji}=\sum_{i=1}^{n}\sum_{j=1}^{n}b_{ij}a_{ji} i=1∑nj=1∑naijbji=i=1∑nj=1∑nbijajitrABtr ABtrAB也可以这样来看：∑axybyxx,y∈1,2,⋯n\sum a_{xy}b_{yx}\quad x,y \in 1,2,\cdots n\quad∑axybyxx,y∈1,2,⋯n其中xyxyxy 有n2n^{2}n2 种组合
同理  trBA=∑bxyayxx,y∈1,2,⋯n\,tr BA=\sum b_{xy}a_{yx}\quad x,y \in 1,2,\cdots ntrBA=∑bxyayxx,y∈1,2,⋯n
可见 xyxyxy 从 111111 取到 nnnnnn，显然两者是等价的

另一个角度理解

因为下标 i,ji,ji,j 互不影响，是可以互换顺序的
trBA=∑i=1n∑j=1nbijaji=∑j=1n∑i=1nbijajitr BA=\sum_{i=1}^{n}\sum_{j=1}^{n}b_{ij}a_{ji}=\sum_{j=1}^{n}\sum_{i=1}^{n}b_{ij}a_{ji} trBA=i=1∑nj=1∑nbijaji=j=1∑ni=1∑nbijaji再用 j′,i′j',i'j′,i′ 替换i,ji,ji,j 得到trBA=∑i′=1n∑j′=1nbj′i′ai′j′=∑i′=1n∑j′=1nai′j′bj′i′tr BA=\sum_{i'=1}^{n}\sum_{j'=1}^{n}b_{j'i'}a_{i'j'}=\sum_{i'=1}^{n}\sum_{j'=1}^{n}a_{i'j'}b_{j'i'}trBA=i′=1∑nj′=1∑nbj′i′ai′j′=i′=1∑nj′=1∑nai′j′bj′i′这时就可以看出来 trAB=trBAtr AB= tr BAtrAB=trBA