Is $V=S\oplus T$ if and only if $S\times T\ni(v,w)\mapsto v+w\in V$ is a vector space isomorphism?

Question

Definition (direct sum): If $S$ and $T$ are vector subspaces of $V$, we write $V=S\oplus T$ if $S\cap T=\{0\}$ and $\{v+w:v\in S,w\in T\}=V$.

Thus, if $V=S\oplus T$, \begin{align} \phi\colon S\times T&\to V\\ (v,w)&\mapsto v+w \end{align} is a vector space isomorphism.

Conversely, can we show that $V=S\oplus T$ if $\phi$ is a vector space isomorphism? That is, can we prove $S\cap T=\{0\}$, possibly without assuming that $V$ is finite dimensional?

Answer 1

Yes, $S\cap T=\{0\}$ if and only if \begin{align} \phi\colon S\times T&\to V\\ (v,w)&\mapsto v+w \end{align} is injective. (Reminder: $S\cap T$ is a vector subspace of $V$, because the intersection of vector subspaces is a vector subspace.)

Proof.

• If $v\in S\cap T$, $\phi(v,0)=\phi(0,v)$ and if, in addition, $\phi$ is injective, \begin{equation} \phi(v,0)=\phi(0,v)\Leftrightarrow (v,0)=(0,v)\Leftrightarrow v=0. \end{equation}
• Conversely, if $S\cap T=\{0\}$ and $(v,w),(v',w')\in S\times T$, then \begin{equation} \phi(v,w)=\phi(v',w')\Leftrightarrow v+w=v'+w'\Leftrightarrow v-v'=w'-w=:u. \end{equation} Since $v-v'\in S$ and $w'-w\in T$, $u\in S\cap T=\{0\}$. Thus, $v=v'$ and $w=w'$.