# Equivalent characterization of inner direct sum of subspaces

Let $V$ be a vector space and let $A, B$ be subspaces of V. The sum of $A, B$ is the subspace of $V$ given by $$A + B = \{a+b : a\in A, b\in B\}$$ Moreover if $A \cap B=\{0\}$ then the sum of $A,B$ is called the inner direct sum of $A,B$ and is denoted with $A \oplus B$. Let $A, B, C$ be subspaces of $V$ with $C = A + B$. Then $$C = A \oplus B \iff \forall c \in C, \exists! a\in A, \exists! b \in B : c = a + b$$

If we replace on the right hand side $\forall c\in C$ with $\exists c\in C$ then does the equivalence still hold?

• Your question does not seem to be clear. Jul 15 '18 at 14:51
• Take $A \cap B = \{0\}$ and $C = A$, then for any $c \in C$ (in particular: $\exists c \in C$ s.t. ...) there are unique $a \in A$ and $b \in B$ s.t. $c = a + b$, namely $a = c$ and $b = 0$, but $C \neq A + B$, hence $C = A \oplus B$ is not true. Jul 15 '18 at 14:54
• @qwertz i restated my quastion because it was not clear enough. Thank u very much for your answer Jul 15 '18 at 15:14
• @nature1729 u are right. I restated my question. Jul 15 '18 at 15:14
• @AnyAD unique. So $\exists !$ means there is EXACTLY ONE Jul 15 '18 at 15:38

The equivalence still holds: We have that \begin{align*} C = A \oplus B &\iff \forall c \in C, \exists! a\in A, \exists! b \in B : c = a + b \\ &\,\implies \exists c \in C, \exists! a\in A, \exists! b \in B : c = a + b \end{align*} because we can choose $c = 0$. To show the other implication $$\exists c \in C, \exists! a\in A, \exists! b \in B : c = a + b \implies \forall c \in C, \exists! a\in A, \exists! b \in B : c = a + b$$ let $c \in C$ such that for $c = a + b$ with $a \in A$, $b \in B$ the summands $a, b$ are unique.

Let $x \in A \cap B$. It then follows from $x \in A$ that $a + x \in A$ and it follows from $x \in B$ that $b - x \in B$. We have that $$c = a + b = (a + x) + (b - x)$$ so it follows from the uniqueness of $a, b$ that $a+x = a$ and $b-x = b$. Both equations show that $x = 0$. This shows that $A \cap B = \{0\}$ and together with $C = A+B$ that $C = A \oplus B$.

For subspaces $A, B, C \subseteq V$ with $C = A + B$ one can show more generally that the following conditions are all equivalent:

• $C = A \oplus B$, i.e. $A \cap B = 0$
• $\forall c \in C: \exists! a \in A \exists! b \in B: c = a + b$
• $\forall c \in C: \exists! a \in A \exists b \in B: c = a + b$
• $\forall c \in C: \exists a \in A \exists! b \in B: c = a + b$
• $\exists! a \in A \exists! b \in B: 0 = a + b$
• $\exists! a \in A \exists b \in B: 0 = a + b$
• $\exists a \in A \exists! b \in B: 0 = a + b$
• You are excellent Jul 15 '18 at 15:47