Problem with direct sums

Question

If $H_1 \oplus H_2 \oplus \cdots \oplus H_k$ and $v_1 \in H_1, v_2 \in H_2 \cdots$ etc.

Why is it that when $v_1 +v_2 + \cdots + v_k=0$

It implies that $v_1 = v_2 = \cdots = v_k = 0$ ?

How does 'unique expression' fit into this?

Answer 1

The summation $v_1+\cdots+v_k$ only makes sense if the $v_i$ all belong to the same module.

So actually $v_i\in H_1\oplus \cdots \oplus H_k$ for $i=1,\dots,k$.

Elements of $H_1\oplus \cdots \oplus H_k$ can be looked at as tuples $(u_1,\dots,u_k)$ and we have the addition:$$(u_1,\dots,u_k)+(w_1,\dots,w_k)=(u_1+w_1,\dots,u_k+w_k)$$

E.g. the expression $v_1\in H_1$ must be read as $(v_1,0,0,\dots,0)\in H_1\oplus \cdots \oplus H_k$.

$H_1$ is identified with submodule $\{(x_1,0,0,\dots,0):x_1\in H_1\}$ but it is important to keep the distinction in mind.

If $\hat{v}_{i}:=\left(0,\dots,0,v_{i},0,\dots,0\right)$ then $\left(v_{1},\dots,v_{k}\right)=\hat{v}_{1}+\cdots+\hat{v}_{k}$. A unique expression.

This with: $$\hat{v}_{1}+\cdots+\hat{v}_{k}=0\iff v_{1}=0\wedge\cdots\wedge v_{k}=0$$

Answer 2

(I suppose that all the $H_i$ are all sub-"foo" of a common "foo" $H$, e.g. "foo" is a vector space, a module, a group, ...)

Yes it implies that $v_1=v_2 =...=v_k=0$.

And this is direct consequence of "unique expression": we have $0\in H_1$, $0\in H_2$, ... and we can write:

$$0 + 0 + 0 + \dots + 0 = 0$$

each "0" in the left member belonging to a $H_i$. This is a decomposition of $0$ and since the sum is direct, this is the only one.

Hence "$v_1+v_2 +\dots +v_k =0$" is the same decomposition a so $v_1=v_2 =\dots=0$.