$\newcommand{\Span}{\mathrm {span}}$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

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

1 Answer 1


$\newcommand{\span}{\rm {span}}$ 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.


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