0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $v=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.