Prove that $S_1+S_2=S_1\oplus S_2$.

Question

Let $\text{dim}V<\infty$ and $S_1,S_2\subset V$ subspaces of $V$. If $S_1\times S_2≈S_1+S_2$, then show that $S_1+S_2=S_1\oplus S_2$.

I think there is a map $T:S_1\times S_2\to S_1+S_2$ defined by $T(s_1,s_2)=s_1+s_2$.

I think there is an use of isomorphism but what is that? How to approach? Also what about if $\text{dim}V=\infty$? Thanks!

Answer 1

In many textbooks, $V=S_1\oplus S_2$ is taken to mean $V=S_1+S_2$ such that $S_1\cap S_2=\{0\}$.

Suppose $S_1\times S_2\approx S_1\oplus S_2$ via some isomorphism $T:(x,y)\mapsto x+y$. Then $T$ clearly maps $S_1\times0$ to $S_1$ (onto and 1-1), and $0\times S_2$ to $S_2$. Since $T$ is invertible, if $v\in S_1\cap S_2$, then $$T^{-1}v\in (S_1\times0)\cap(0\times S_2)=\{(0,0)\}\implies v=0$$

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$\color{gray}{\textrm{Edit: Another proof that is valid only for finite dimensions but where $T$ can be any}}$ $\color{gray}{\textrm{isomorphism is as follows:}}$ \begin{align} \color{gray}{\dim(S_1+S_2)}&\color{gray}{=\dim(S_1)+\dim(S_2)-\dim(S_1\cap S_2)}\\ \color{gray}{\dim(S_1\times S_2)}&\color{gray}{=\dim(S_1)+\dim(S_2)} \end{align} $\color{gray}{\textrm{So if $S_1\times S_2\approx S_1+S_2$, then $\dim(S_1\cap S_2)=0$ and thus $S_1\cap S_2=\{0\}$.}}$

$\color{gray}{\textrm{However this is not true in infinite dimensions. For example, consider the vector space}}$ $\color{gray}{\textrm{of real sequences}}$ $\color{gray}{X=\mathbb{R}^\mathbb{N}=\mathbb{R}\times\mathbb{R}\times\cdots}$, $\color{gray}{\textrm{let }}$ $\color{gray}{S_1:=\{(x,0,0,\ldots):x\in\mathbb{R}\}}$, $\color{gray}{S_2:=X}$. $\color{gray}{\textrm{Then $S_1+S_2=X\approx S_1\times S_2$ via the map $(x_1,x_2,\ldots)\mapsto(x_1,(x_2,\ldots))$,}}$ $\color{gray}{\textrm{yet}}$ $\color{gray}{S_1\cap S_2=S_1\ne\{0\}}$.

Answer 2

Let $S_1, S_2 \subseteq V$ and $S_1 + S_2 = \{s_1+s_2\mid s_i\in S_i,\, i = 1,2\}.$ Then the following are equivalent:

1. For all $v\in S_1+S_2$, there exist unique $s_i\in S_i,\, i =1,2$, such that $v = s_1 + s_2$,
2. $S_1 \cap S_2 = \{0\}.$

The proof is standard and should be somewhere in your textbook. It's a good exercise, anyway.

Now, note that $\dim(S_1+S_2) = \dim S_1 + \dim S_2 - \dim(S_1\cap S_2)$ and $\dim(S_1\times S_2) = \dim S_1 + \dim S_2$. From this, it should be clear that $\dim(S_1\cap S_2) = 0$ if and only if $\dim(S_1+S_2) = \dim (S_1\times S_2)$ if and only if $S_1+ S_2$ and $S_1\times S_2$ are isomorphic.

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Note that in the above we didn't need to suppose that isomorphism between $S_1\times S_2$ is of the form $T(s_1,s_2) = s_1+s_2$, but let me say something about this map. If you look at it closely, $S_1 + S_2$ is defined to be the image of $T$, so $T$ is epimorphism by construction. Now, note that the property 1. is equivalent to $T$ being injective. So, in finite dimensional case, if there is any isomorphism between $S_1+ S_2$ and $S_1\times S_2$, then $T$ is isomorphism. This may not be true in infinite dimensional case.

Let $V$ be a vector space with basis $\{e_n\}_{n\in\mathbb N}$. For example, the space of all finite real sequences, in which case $e_n$ is a sequence whose $n$-th term is $1$, and all other terms are $0$. Define $S_1 = \operatorname{span}\{e_1\}$ and $S_2 = V$. We have $S_1 + S_2 = V$ and $S_1\cap S_2 = S_1$. On the other hand, $S_1\times S_2$ has basis $\{(e_1,0), (0,e_1),(0,e_2),\ldots\}$. Now define linear map $A\colon S_1\times S_2\to S_1+S_2$ on the basis vectors such that $(e_1,0)\mapsto e_1$ and $(0,e_n)\mapsto e_{n+1}$, for all $n\in\mathbb N$. This map is an isomorphism and thus we found counterexample to the claim in case of an infinite dimensional vector space.