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 $0$.

and the proposition 3.16:

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 $v=0$.


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