# Decomposing linear mapping between direct sums of vector spaces

In a textbook I found the following exercise:

Let $V = V_1 \oplus V_2 \oplus \cdots \oplus V_n$ and $W = W_1 \oplus W_2 \oplus \ldots \oplus W_n$ and $A_i \in \mbox{hom}(V_i, W_i)$ for $i = 1,\ldots n$. Show that there exists a unique transformation $A : V \to W$ such that $$A(v_1 + v_2 + \ldots + v_n) = A_1v_1 + A_2v_2 + \ldots + A_nv_n.$$ for $v_i \in V$. For me the solution is quite obvious, suppose we have two such transformations $A,B \in \mbox{hom}(V,W)$. For some $v \in V$ we have a unique decomposition $v = v_1 + v_2 + \ldots + v_n$ as a property of the direct sum, so that $$B(v) = B(v_1 + v_2 + \ldots + v_n) = A_1 v_1 + A_2 v_2 + \ldots + A_n v_n = A(v_1 + v_2 + \ldots + v_n) = Av$$ implying $Bv = Av$ for every $v \in V$, showing that $B = A$. Now I asked myself if every $A \in \mbox{hom}(V,W)$ could be decomposed in such $A_i \in \mbox{hom}(V_i, W_i)$, i.e. for every $A \in \mbox{hom}(V,W)$ there exists $A_i : V_i \to W_i$ such that $$A(v_1 + v_2 + \ldots + v_n) = A_1v_1 + A_2v_2 + \ldots + A_nv_n$$ for $v_i \in V_i$. I guess, as every $v_i \in V_i$ corresponds to a unique vector $v_i \in V$, we can define $$A_iv_i := P_i Av_i$$ where $P_i : V \to V_i$ denotes the unique projection onto the $i$-the space $V_i$. This gives a mapping $A_i : V_i \to W_i$. Then $$A(v_1 + v_2 + \ldots + v_n) = Av_1 + Av_2 + \ldots + Av_n$$ but $Av_i \in W$ and here I am stuck. So my question, is such an decomposition possible and am I on the right track?

Two things:

1. You proved the uniqueness, but not the existence. You have to check that the putative map is a linear transformation.

2. No, it's not true that all $A$ arise this way. Think of a two-dimensional vector space which is a direct sum of two one-dimensional vector spaces. Then the set of all $A$ is the set of all $2\times2$ matrices, but the $A$ which arise as sums of are the diagonal matrices.

• Thanks for pointing out, I forgot existence. The map $A$ defined that way is actually linear, for if $u,v$ with $v = v_1 + \ldots v_n$ and $u = u_1 + \ldots + u_n$ we have $A(u+v) = A((u_1+v_1)+\ldots+(u_n+v_n)) = A_1(u_1+v_1) + \ldots + A_n(u_n+v_n) = A_1u_1 + \ldots + A_nu_n + A_1 v_1 + \ldots + A_n v_n$, where commutativity and linearity of the $A_i$'s was used in the rearrangements. The fact that $A(\lambda v) = \lambda A(v)$ is shown similar. Sep 12, 2014 at 12:47

Take $V_1 = V_2 = \mathbb{R}$ and also $W_1 = W_2 = \mathbb{R}$. Let the basis vectors be $v_1,v_w,w_1,w_2$ respectively.

A simple counterexample is $A : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $A(v_1) = w_2$ and $A(v_2)=w_1$.

Note that $A$ maps $V_1$ entirely onto $W_2$, and $V_2$ entirely onto $W_1$. This would be the "failing point" of your proof - where the compositions with the projection would simply be the zero map in each case, and not recombine to make the composite map $A$.