product summation map — Jemoka Knowledge Base

Let U_1, \dots, U_{m} be subspaces of V; we define a linear We define \Gamma to be a map U_1 \times \dots U_{m} \to U_1 + \dots + U_{m} such that:

\begin{equation} \Gamma (u_1, \dots, u_{m}) = u_1 + \dots + u_{m} \end{equation}

Essentially, \Gamma is the sum operation of the elements of the tuple made by the Product of Vector Spaces. U_1 + \dots + U_{m} is a direct sum IFF \Gamma is injective Proof: Given \Gamma is injective: Given injectivity, we have that injectivity implies that null space is \{0\}. Now, because the only way to produce 0 is to have the input product/tuple be 0, u_1 \dots u_{m} = 0. So, given a sum of subsets is a direct sum IFF there is only one way to write 0, the sum is a direct sum. Given direct sum: Reverse the logic of above directly. Given its a direct sum, the only way to be in the null space of \Gamma (i.e. have the sum of the elements of tuple by 0) is by taking each u_1 \dots u_{m} to 0. Now, injectivity implies that null space is \{0\}, so \Gamma is injective. \blacksquare Aside: \Gamma is surjective because product of vector-spaces is simply the pre-combined version of the sum. So a corollary of the above result is that: U_1 + \dots + U_{m} is a direct sum IFF \Gamma is invertable, because injectivity and surjectivity implies invertability. U_1 + \dots + U_{m} is a direct sum IFF \dim (U_1 + \dots + U_{m}) = \dim U_1 + \dots + \dim U_{m} \Gamma is surjective for all cases because product of vector-spaces is simply the pre-combined version of the sum. So, by rank-nullity theorem, \dim (U_1 \times \dots U_{m}) = \dim null\ \Gamma + \dim (U_1 + \dots + U_{m}). Now, \dim null\ \Gamma = 0 IFF \dim (U_1 \times \dots U_{m}) = 0 + \dim (U_1 + \dots + U_{m}). Now, dimension of the Product of Vector Spaces is the sum of the spaces’ dimension. So: \dim null\ \Gamma = 0 IFF \dim U_1 + \dots + \dim U_{m} = 0 + \dim (U_1 + \dots + U_{m}). Now, U_1 + \dots + U_{m} is a direct sum IFF \Gamma is injective, and from above \dim null\ \Gamma = 0 (that \Gamma is injective) IFF \dim U_1 + \dots + \dim U_{m} = 0 + \dim (U_1 + \dots + U_{m}). So, U_1 + \dots + U_{m} is a direct sum IFF \dim (U_1 + \dots + U_{m}) = \dim U_1 + \dots + \dim U_{m}, as desired. \blacksquare (Note that this proof is built out of a series of IFFs, so it goes in both directions.)