矩阵B与A交换 则B与adj(A)交换

hbghlyj

已知$\DeclareMathOperator{\adj}{adj}AB=BA$, 求证$\adj(A)B=B\adj(A)$.
当$A$可逆,$A^{-1}B=A^{-1}BAA^{-1}=A^{-1}ABA^{-1}=BA^{-1}$.
当$A$不可逆

hbghlyj

det(A+t I) is a polynomial in t with degree at most n, so it has at most n roots. Note that the ij th entry of adj((A+t I)(B)) is a polynomial of at most order n, and likewise for adj(A+t I) adj(B). These two polynomials at the ij th entry agree on at least n + 1 points, as we have at least n + 1 elements of the field where A+t I is invertible, and we have proven the identity for invertible matrices. Polynomials of degree n which agree on n + 1 points must be identical (subtract them from each other and you have n + 1 roots for a polynomial of degree at most n – a contradiction unless their difference is identically zero). As the two polynomials are identical, they take the same value for every value of t. Thus, they take the same value when t = 0.

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Source: Adjugate matrix#Properties

hbghlyj

Suppose that A commutes with B. Multiplying the identity AB = BA on the left and right by adj(A) proves that $ \det(\mathbf {A} )\operatorname {adj} (\mathbf {A} )\mathbf {B} =\det(\mathbf {A} )\mathbf {B} \operatorname {adj} (\mathbf {A} )$. If A is invertible, this implies that adj(A) also commutes with B.  Over the real or complex numbers, continuity implies that adj(A) commutes with B even when A is not invertible.

红色部分详细步骤:

AB = BA

左乘和右乘 adj(A) 得

adj(A) AB adj(A) = adj(A) BA adj(A)

由 adj(A) A = A adj(A) = det(A) I 得

det(A) B adj(A) = det(A) adj(A) B

Czhang271828

实际上可交换矩阵的可交换矩阵等价于原矩阵的多项式.

可以直接猜测 $\mathrm{Adj}(A)$ 是 $A$ 的多项式. 记 $\chi$ 是特征多项式, $A_x:=xI-A$, 则当 $x\notin \mathrm{spec}(A)$ 时有
\[
A_x\cdot \mathrm{adj}(A_x)=\chi_A(x)\cdot I.
\]
此时 $\chi_{A_x}(0)\neq 0$, 故多项式 $\chi_A(x)\cdot \left.\dfrac{[\chi_{A_x}(0)^{-1}\chi_{A_x}(\lambda)-1]}{\lambda}\right|_{\lambda=A_x}=\mathrm{adj}(A_x)$. 再令 $x=0$ 即可.