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Answers for “如果 A 是 m 乘 n,B 是 p 乘 q,为什么 A \otimes B 的大小等于 m p 乘 n q?”

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The Kronecker product A⊗BA \otimes B is an mp×nqmp \times nq block matrix because it replaces each element aija_{ij} of the m×nm \times n matrix AA with the p×qp \times q matrix aijBa_{ij}B. This results in mm row blocks and nn column blocks, each of size p×qp \times q, making the total dimensions mp×nqmp \times nq.

Conditions: AA is an m×nm \times n matrix.; BB is a p×qp \times q matrix.; aija_{ij} is an element of AA.

Understand why

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The isomorphism hom⁡(U,hom⁡(V,W))≅hom⁡(U⊗V,W)\hom(U, \hom(V, W)) \cong \hom(U \otimes V, W) reflects the adjoint/duality relationship between Hom and tensor product. A linear map from UU to the space of linear maps from VV to WW corresponds to a bilinear map from U×VU \times V to WW, which in turn corresponds to a linear map from the tensor product space U⊗VU \otimes V to WW.

Conditions: U,V,WU, V, W are linear spaces over the number field P\mathbb{P}.; hom⁡(X,Y)\hom(X,Y) denotes the vector space of all linear maps from XX to YY.

Understand why

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The tensor product notation u⊗vu \otimes v satisfies distributive laws and scalar mobility rules because it is constructed via a bilinear map φ\varphi. Bilinearity implies linearity in each variable separately, leading to (a1u1+a2u2)⊗v=a1u1⊗v+a2u2⊗v(a_1u_1+a_2u_2)\otimes v=a_1u_1\otimes v+a_2u_2\otimes v, u⊗(b1v1+b2v2)=b1u⊗v1+b2u⊗v2u\otimes(b_1v_1+b_2v_2)=b_1u\otimes v_1+b_2u\otimes v_2, and a(u⊗v)=(au)⊗v=u⊗(av)a(u\otimes v)=(au)\otimes v=u\otimes(av).

Conditions: u,u1,u2u, u_1, u_2 belong to UU.; v,v1,v2v, v_1, v_2 belong to VV.; a,a1,a2,b1,b2a, a_1, a_2, b_1, b_2 are scalars in the field P\mathbb{P}.