---
title: "product summation map"
source: https://www.jemoka.com/posts/kbhproduct_summation_map/
---

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&rsquo; 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.)
