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So the question is the following: Prove the following are equivalent:

  1. $V$ is the inner direct sum of the subspaces $U_1,U_2,\dots ,U_n$.

  2. $V = U_1+\dots+U_n$ and $\dim(V) = \dim(U_1)+\dots+\dim(U_n)$.

  3. Every vector $v$ in $V$ can be written as $v=u_1+u_2+\dots+u_n$, which $u_i$ belong to $U_i$ in a unique way.

I'stuck at 2 implies 3 and 3 implies 1 .

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  • $\begingroup$ I had tried 1)implies 2) $\endgroup$
    – EK1222
    Apr 3 '15 at 20:28
  • $\begingroup$ @EKIMKAM What is your definition of inner direct sum? $\endgroup$
    – Eoin
    Apr 3 '15 at 20:28
  • $\begingroup$ 3 implies 1 can be proven by using the fact that the combination is unique. Take for example $u \in (U_{1}\cup...\cup U_{i})\cap U_{i+1}$. Note that we can write $u$ in two ways: $u=u_{1}+...+u_{i}+0$ and $u=0+...+0+u_{i+1}$. $\endgroup$
    – eranreches
    Apr 3 '15 at 20:31
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Let's do $2) \implies 3)$. We are given: $V = U_1 + \cdots + U_n$ and:

$\dim(V) = \dim(U_1) + \cdots + \dim(U_n)$.

Suppose $\dim(U_k) = m_k$ ,and $\dim(V) = m$, so that $m = m_1 + \cdots +m_n$.

Let $\{u_{k1},\dots,u_{km_k}\}$ be a basis for $U_k$.

Then $\{u_{11},\dots,u_{1m_1},u_{21},\dots,u_{2m_2},\dots,u_{n1},\dots,u_{nm_n}\}$ spans $V$, since

$V = U_1 + \cdots + U_n$. Since this set has $m_1 + \cdots + m_n = m = \dim(V)$ elements, it is a basis for $V$ (it is a minimal spanning set, since any subset has $< m = \dim(V)$ elements).

So suppose $v = u_1 + \cdots + u_n = u_1' + \cdots + u_n'$, with $u_k,u_k' \in U_k$.

Then $0 = v - v = (c_{11} - c_{11}')u_{11} + \cdots (c_{1m_1} - c_{1m_1}')u_{1m_1} + \cdots + (c_{n1} - c_{n1}')u_{n1} + \cdots + (c_{nm_n}-c_{nm_n}')u_{nm_n}$

for some scalars $c_{kj_k},c_{kj_k}'$ ($k$ runs from $1$ to $n$, and $j_k$ runs from $1$ to $m_k$).

By the linear independence of the $\{u_{kj_k}\}$ (which form a basis for $V$), we have that $c_{kj_k} = c_{kj_k}'$

for every $k$ and $j_k$, so that in particular:

$u_k = c_{k1}u_{k1} +\cdots + c_{km_k}u_{km_k} = c_{k1}'u_{k1} + \cdots +c_{km_k}'u_{km_k} = u_k'$, for eack $k$, so the $u_k$ are unique.

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