${\rm End}(\bigoplus V)=\bigoplus({\rm End}(V))$

Question

I saw in a demonstration that they used the fact $${\rm End}(\bigoplus V)=\bigoplus({\rm End}(V))$$ for $\bigoplus$ finite and $V$ a space vectorial. He tried to prove it but I don't have a clear idea. Could you give me a suggestion for this proof?

Answer 1

Use additivity of Hom, and note the sums go over different index sets $$\begin{align} {\rm End}\left(\bigoplus_{i=1}^n V\right) &= {\rm Hom}\left(\bigoplus_{i=1}^n V, \bigoplus_{i=1}^n V\right) \\ &= \bigoplus_{i=1}^n \bigoplus_{i=1}^n {\rm End}(V) \\ &= \bigoplus_{1\leq i,j\leq n} {\rm End}(V). \end{align}$$

Answer 2

You can think of $\operatorname{End}(V_1 \oplus \cdots \oplus V_n)$ as matrices $(T_{i,j})_{i,j = 1}^n$ where $T_{i,j} : V_i \to V_j$.

So if all the $V_i$'s are 1-dimensional then $T_{i,j}$ is represented by a scalar and hence $(T_{i,j})$ is just an $n \times n$ matrix. This makes sense based on what we know: $\operatorname{End}(\mathbf{F}^n) = \operatorname{Mat}_{n}(\mathbf F)$.

More generally, we represent $\operatorname{End}(V_1 \oplus \cdots \oplus V_n)$ as block matrices of the appropriate shape.