Doubt about the notion of "unique representation of a vector $v$" given by Direct Sums and Basis sets

$$\newcommand{\Span}{\mathrm {span}}$$Consider only finite dimensional vector spaces with a "ordinary" field (like $$\mathbb{R}$$ or $$\mathbb{C}$$). Moreover I appreciate a discussion based only on elementary Linear Algebra like exposed on [1] or [2].

So, you have then Vector spaces, subspaces, the $$\Span(W)$$ set, the notion of a Linear Independent set of vectors (and, conversely, the notion of Linear dependent set of vectors).

With these notions (and elementary set theory) we can construct some other vector structures:

$$1)$$ The Sum of two subspaces: $$U + W$$

$$2)$$ The Direct Sum of two subspaces $$U \oplus W$$

$$3)$$ A basis for a vector space.

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Now, a basis for a vector space $$V$$ is a set $$\mathcal{B} = \{u_{1},...,u_{n}\}$$ which are linear independent and spans the entire $$V$$. There's a theorem which says that given a basis for $$V$$, each vector of $$V$$ is written uniquely as a linear combination of the elements of $$\mathcal{B}$$:

$$v = a_{1}u_{1}+...+a_{n}u_{n} \tag{1}$$

That's because $$v\in V$$, the $$\Span(\mathcal{B}) = V$$ and the set of vectors $$\mathcal{B}$$ is linear independent, so the scalars $$a_{1},...,a_{n}$$ are defined for each element of $$V$$.

Meanwhile, there's another notion of "each vector of $$V$$ is written uniquely", this notion is given by vectors of direct sums. Now, my professors wrote an interresting equation, and he said that the following equation have the same notion as $$(1)$$. This equation is:

$$v \in \Span(\{u_{1}\}) \oplus ... \oplus \Span(\{u_{n}\}) \tag{2}$$

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My question is: why $$(2)$$ is equivalent to $$(1)$$, concerning the notion of "an vector uniquely determined" ?

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[1] HOFFMAN & KUNZE - Linear Algebra

[2] PAUL HALMOS - Finite Dimensional Vector Spaces

• Are you sure equation (2) shouldn't have the symbol $\in$ rather than $=$? Commented Sep 6, 2019 at 1:24

$$\newcommand{\span}{\rm {span}}$$ Suppose $$V$$ is a vector space and $$V=\span\{u_1,u_2\}$$($$u_1$$ and $$u_2$$ are assumed to be linearly independent). Consider the subspaces $$\span\{u_1\}$$ and $$\span\{u_2\}.$$ Suppose $$v\in \span\{u_1\}\cap\span\{u_2\}$$. Then $$v=a_1u_1=a_2u_2\implies a_1u_1-a_2u_2=0.$$ Since $$u_1$$ and $$u_2$$ are linearly independent, $$a_1,a_2=0$$ and hence $$v=0$$. So $$\span\{u_1\}\cap\span\{u_2\}=\{0\}$$ and $$V=\span\{u_1\}+\span\{u_2\}$$. So what can you conclude now?
Conversely, suppose $$V=\span\{u_1\}\oplus\span\{u_2\}$$. Then any element can be written as a sum of elements $$a_1u_1$$ and $$a_2u_2$$ uniquely(here $$a_i$$ is a scalar). In particular, we can write $$0=0u_1+0u_2$$ and since it is unique, we can conclude that $$u_1$$ and $$u_2$$ are linearly independent.