# Alternative proof for “linear map from product of subspaces to a sum of subspaces is injective if and only if the sum is a direct sum”

For the theorem 3.77 in Axler's linear algebra done right:

Suppose that $$U_1, U_2, \ldots U_m$$ are subspaces of $$V$$. Define a linear map $$\Gamma: U_1 \times U_2 \times U_3 \ldots U_m \to U_1 + U_2 > \ldots + U_m$$ by $$\Gamma(u_1, u_2, \ldots u_m) = u_1 + u_2 + \ldots > u_m$$ Then $$U_1 + U_2 + \ldots + U_m$$ is a direct sum if and only if $$\Gamma$$ is injective.

Is this a valid proof?

## Proof that injective $$\Gamma$$ implies direct sum of subspaces

### $$\Gamma$$ is also surjective

For each $$u_1 + u_2 + \ldots u_m \in U_1 + U_2 \ldots U_m \ \exists (u_1, u_2, \ldots, u_m) \in U_1 \times U_2 \times U_3 \ldots U_m$$. So $$\Gamma$$ is surjective.

### $$\Gamma$$ is invertible

$$\Gamma$$ is both surjective and injective, so it is invertible.

### Isomorphism of $$U_1 \times U_2 \times U_3 \ldots U_m$$ and $$U_1 + U_2 \ldots + U_m$$

Because $$\Gamma$$ is invertible, this is equivalent to saying these two spaces are isomorphic. But then, two finite dimensional spaces are invertible if and only if their dimensions match. The product of subspaces has a dimension $$\sum_{i} \text{dim } U_i$$. This is only possible for the sum if and only if $$\text{dim } U_1 \cap U_2 \ldots \cap U_m = 0 \implies U_1 \cap U_2 \ldots \cap U_m = \{ 0 \}$$. This implies that the sum of these subspaces is a direct sum. (this is the part I feel unsure about. Since Axler mentions a proof only for two subspaces, I'm not entirely sure if this is true. But I can see how it can be generalized by using $$U_1 + U_2, U_3$$ as the two subspaces and then imply the same for $$m$$ subspaces.)

## Proof that the sum of subspaces is a direct sum implies injective $$\Gamma$$

I would start again with the dimensional implication. Since the dimension of the intersection of subspaces is zero, this means dimension of the sum of subspaces is $$m$$. This is equal to the dimension of the product of subspaces. This implies an isomorphism exists. Since $$\Gamma$$ is surjective and invertible, this implies it must also be injective.

No, the condition $$U_1 \cap \cdots \cap U_m = \{0\}$$ does not imply that $$U_1 + \cdots + U_m$$ is direct, and, as you said, this only works when $$m=2$$. For example, in $$\mathbb R^2$$ take $$U_1 = \{(x,0) : x \in \mathbb R\}$$, $$U_2 = \{(0,y) : y \in \mathbb R\}$$ and $$U_3 = \{(x,y) \in \mathbb R^2 : x=y\}$$. Clearly $$U_1 \cap U_2 \cap U_3 = \{0\}$$ but $$\mathbb R^2 = U_1 + U_2 + U_3$$ is not direct since there are no a unique way of writting $$(1,1)$$ as a sum $$u_1+u_2+u_3$$ with $$u_i \in U_i$$ for $$i=1,2,3$$: $$(1,1) = (1,0)+(0,1)+(0,0) = (0,0)+(0,0)+(1,1).$$ I suggest you use the proposition 1.44:
Suppose $$U_1,\dots,U_m$$ are subspaces of $$V$$. Then $$U_1 + \cdots + U_m$$ is direct if and only if the only way to write $$0$$ as a sum $$u_1+\cdots+u_m$$, where each $$u_j$$ is in $$U_j$$, is by taking each $$u_j$$ equal to $$0$$.
Let $$T \in \mathcal L(V,W)$$. Then $$T$$ is injective if and only if $$\operatorname{null} T = \{0\}$$.
Observe that $$\operatorname{null} T = \{0\}$$ is equivalent to saying that for each $$v \in V$$ the following holds: if $$Tv=0$$ then $$v=0$$.