"Direct sum" of linear maps

1.9k Views Asked by At

Let $\{V_i\}_{i\in I}$ be a family of vector spaces, and let $A_i\in End(V_i)=Hom(V_i,V_i)$. Show that there is a unique element $\bigoplus_{i\in I}A_i\in End(\bigoplus_{i\in I}V_i)$ whose restriction to the image of $V_i$ in the sum is $A_i$.

My attempt:

Uniqueness: Suppose $\overline{\bigoplus_{i\in I}A_i}$ is another linear map from direct sum of $V_i$'s to itself which agrees with $\bigoplus_{i\in I}A_i$ on $A_i$'s. For every $v\in\bigoplus_{i\in I}V_i$, $\overline{\bigoplus_{i\in I}A_i}(v)=\overline{\bigoplus_{i\in I}A_i}(\sum_{j\in B}\alpha_jb_j)=\sum_{j\in B}\alpha_j\overline{\bigoplus_{i\in I}A_i}(b_j)$ where $b_j$ are basis of $A_i$ for some $i$. This is the same for $\bigoplus_{i\in I}A_i$. So $\overline{\bigoplus_{i\in I}A_i}(v)=\bigoplus_{i\in I}A_i$ everywhere proved uniqueness.

How to show existence?

1

There are 1 best solutions below

0
On

Your proof for uniqueness looks good, but the notation could use some touching up. Since we haven't yet shown that $\bigoplus_{i\in I} A_i$ is unique, it's a little presumptuous to use that symbol. After all, we're not yet sure it's unambiguously defined. (Plus, then you can use maps like $L$ and $L'$, which are much easier to read :) )

Existence: Many (most?) existence proofs can be done by just constructing the object you want, and handing it to the reader. Since $\bigoplus$ can be realized as ordered tuples (with finitely many non-zero components, define $L \in End(\bigoplus_i V_i)$ to be: $$ L(x_1, \ldots, x_\alpha) = (A_1(x_1), \ldots, A_\alpha(x_\alpha)) $$