矩阵的迹是特征值的加和，也即矩阵A的主对角线元素的总和。

对于方阵 A = [ a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a n 1 ⋯ a n n ] A=\left[\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n n}\end{array}\right] A=⎣⎢⎡​a11​⋮an1​​⋯⋱⋯​a1n​⋮ann​​⎦⎥⎤​

tr ⁡ A = ∑ i = 1 n a i i \operatorname{tr} A=\sum_{i=1}^{n} a_{i i} trA=i=1∑n​aii​

① trA ⁡ = t r A T \operatorname{trA}=t r A^{T} trA=trAT

证明：

A = [ a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a n 1 ⋯ a n n ] A=\left[\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n n}\end{array}\right] A=⎣⎢⎡​a11​⋮an1​​⋯⋱⋯​a1n​⋮ann​​⎦⎥⎤​ ​ A T = [ a 11 ⋯ a n 1 ⋮ ⋱ ⋮ a 1 n ⋯ a n n ] A^T=\left[\begin{array}{ccc}a_{11} & \cdots & a_{n 1} \\ \vdots & \ddots & \vdots \\ a_{1 n} & \cdots & a_{n n}\end{array}\right] AT=⎣⎢⎡​a11​⋮a1n​​⋯⋱⋯​an1​⋮ann​​⎦⎥⎤​

主对角元素不变，显然 trA ⁡ = t r A T \operatorname{trA}=t r A^{T} trA=trAT

② tr ⁡ A B = tr ⁡ B A \operatorname{tr} A B=\operatorname{tr} B A trAB=trBA

证明 tr ⁡ A B = tr ⁡ B A \operatorname{tr} A B=\operatorname{tr} B A trAB=trBA

假设有矩阵 A m × n , B n × m A_{m \times n}, B_{n \times m} Am×n​,Bn×m​, 则 C m × m = A m × n B n × m , D n × n = B n × m A m × n C_{m \times m}=A_{m \times n} B_{n \times m}, \quad D_{n \times n}=B_{n \times m} A_{m \times n} Cm×m​=Am×n​Bn×m​,Dn×n​=Bn×m​Am×n​

tr ⁡ ( A B ) = tr ⁡ ( C ) = ∑ i = 1 m c i i = ∑ i = 1 m ∑ j = 1 n a i j b j i tr ⁡ ( B A ) = tr ⁡ ( D ) = ∑ j = 1 n d j j = ∑ j = 1 n ∑ i = 1 m b j i a i j ∴ tr ⁡ ( A B ) = tr ⁡ ( B A ) \begin{aligned} &\operatorname{tr}(A B)=\operatorname{tr}(C)=\sum_{i=1}^{m} c_{i i}=\sum_{i=1}^{m} \sum_{j=1}^{n} a_{i j} b_{j i} \\ &\operatorname{tr}(B A)=\operatorname{tr}(D)=\sum_{j=1}^{n} d_{j j}=\sum_{j=1}^{n} \sum_{i=1}^{m} b_{j i} a_{i j} \\ &\therefore \operatorname{tr}(A B)=\operatorname{tr}(B A) \end{aligned} ​tr(AB)=tr(C)=i=1∑m​cii​=i=1∑m​j=1∑n​aij​bji​tr(BA)=tr(D)=j=1∑n​djj​=j=1∑n​i=1∑m​bji​aij​∴tr(AB)=tr(BA)​

推论： tr ⁡ A B C = tr ⁡ C A B = tr ⁡ B C A \operatorname{tr} A B C=\operatorname{tr} C A B=\operatorname{tr} B C A trABC=trCAB=trBCA， tr ⁡ A B C D = tr ⁡ D A B C = tr ⁡ C D A B = tr ⁡ B C D A \operatorname{tr} A B C D=\operatorname{tr} D A B C=\operatorname{tr} C D A B=\operatorname{tr} B C D A trABCD=trDABC=trCDAB=trBCDA

③ t r a A = a t r A tra A=a trA traA=atrA

证明：

A = [ a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a n 1 ⋯ a n n ] A=\left[\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n n}\end{array}\right] A=⎣⎢⎡​a11​⋮an1​​⋯⋱⋯​a1n​⋮ann​​⎦⎥⎤​

a A = a [ a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a n 1 ⋯ a n n ] aA=a\left[\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n n}\end{array}\right] aA=a⎣⎢⎡​a11​⋮an1​​⋯⋱⋯​a1n​⋮ann​​⎦⎥⎤​

t r a A traA traA= Σ i = 1 n Σ j = 1 n a ∗ a i a j = a Σ i = 1 n Σ j = 1 n a i a j \Sigma_{i=1}^n\Sigma_{j=1}^na*a_ia_j=a\Sigma_{i=1}^n\Sigma_{j=1}^na_ia_j Σi=1n​Σj=1n​a∗ai​aj​=aΣi=1n​Σj=1n​ai​aj​

a t r A atrA atrA= a Σ i = 1 n Σ j = 1 n a i a j a\Sigma_{i=1}^n\Sigma_{j=1}^na_ia_j aΣi=1n​Σj=1n​ai​aj​

故得证

④ ∇ A tr ⁡ A B = B T \nabla_{A} \operatorname{tr} A B=B^{T} ∇A​trAB=BT

证明：

令 A = [ a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a n 1 ⋯ a n n ] A=\left[\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n n}\end{array}\right] A=⎣⎢⎡​a11​⋮an1​​⋯⋱⋯​a1n​⋮ann​​⎦⎥⎤​

trAB ⁡ = ∑ i = 1 n ∑ j = 1 n a i j b j i ∂ Tr ⁡ ( A B ) ∂ A = ∂ ∑ i = 1 n ∑ j = 1 n a i j b j i ∂ ∑ i = 1 n ∑ j = 1 n a i j = ∑ i = 1 n ∑ j = 1 n b j i = B T \operatorname{trAB}=\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i j} b_{j i}\\ \frac{\partial \operatorname{Tr}(A B)}{\partial A}=\frac{\partial \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i j} b_{j i}}{\partial \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i j}}=\sum_{i=1}^{n} \sum_{j=1}^{n} b_{j i}=B^{T} trAB=i=1∑n​j=1∑n​aij​bji​∂A∂Tr(AB)​=∂∑i=1n​∑j=1n​aij​∂∑i=1n​∑j=1n​aij​bji​​=i=1∑n​j=1∑n​bji​=BT

∴ ∇ A tr ⁡ A B = [ b 11 ⋯ b n 1 ⋮ ⋱ ⋮ b 1 n ⋯ b n n ] = B T \therefore \nabla_{A} \operatorname{tr} A B=\left[\begin{array}{ccc}b_{11} & \cdots & b_{n 1} \\ \vdots & \ddots & \vdots \\ b_{1 n} & \cdots & b_{n n}\end{array}\right]=B^{T} ∴∇A​trAB=⎣⎢⎡​b11​⋮b1n​​⋯⋱⋯​bn1​⋮bnn​​⎦⎥⎤​=BT 。 得证

⑤ ∇ A tr ⁡ A B A T C = C A B + C T A B T \nabla_{A} \operatorname{tr} A B A^{T} C=C A B+C^{T} A B^{T} ∇A​trABATC=CAB+CTABT

证明：

∇ A tr ⁡ A B A T C = ∇ A tr ⁡ A ( B A T C ) + ∇ A t r ( C A B ) A T \nabla_{A} \operatorname{tr} A B A^{T} C=\nabla_{A} \operatorname{tr} A\left(B A^{T} C\right)+\nabla_{A} t r(\mathrm{CAB}) \mathrm{A}^{T} ∇A​trABATC=∇A​trA(BATC)+∇A​tr(CAB)AT ​ = ( B A τ C ) T + ∇ A t r [ ( C A B ) A T ] T =\left(\mathrm{BA}^{\tau} \mathrm{C}\right)^{T}+\nabla_{A} t r\left[(\mathrm{CAB}) \mathrm{A}^{T}\right]^{T} =(BAτC)T+∇A​tr[(CAB)AT]T ​ = C J A B τ + ∇ A tr ⁡ A ( C A B ) T =C^{J} A B^{\tau}+\nabla_{A} \operatorname{tr} A(\mathrm{CAB})^{T} =CJABτ+∇A​trA(CAB)T ​ = C τ A B τ + C A B =C^{\tau} A B^{\tau}+C A B =CτABτ+CAB