Direct Sum of $n$ Subspaces

Question

I just need some guidance to prove a portion of the following theorem.

Let $V_1, V_2, ... , V_n$ be subspaces of a vector space $V$. Then the following statements are equivalent.

1. $W = \sum V_i$ is a direct sum.
2. Decomposition of the zero vector is unique.
3. $V_i\cap\sum_{j\neq i}V_i = \{0\}$ for $i = 1, 2, ..., n$
4. dim$W$ = $\sum$dim$V_i$

I know there are many ways to go about proving this. One of which requiresI prove $1 \leftrightarrow 2$, $1 \leftrightarrow 3$, $1 \leftrightarrow 4$. Another requires I prove $1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1$.

I have been able to show that $1 \rightarrow 2, 3 \rightarrow 4$, and $4 \rightarrow 1$. I am having a hard time proving $2 \rightarrow 3$.

To give you an idea of what I have for $2 \rightarrow 3$:

$2.$ states, there is a unique decomposition for $0 = \alpha_1 + \alpha_2 +... \alpha_n$ where $\alpha_i \in V_i$ for $i = 1, 2, ..., n$. Then we can say $-\alpha_i = \alpha_1 + ... + \alpha_{i-1} + \alpha_{i+1} + ... + \alpha_{n} \in V_i$ By assumption, we deduce that $\alpha_i = 0$ for all $i$.

What I have above does not convince me. I have yet to show each subspace is pairwise disjoint; sadly, I am not sure what to do from here and what I am thinking of might be incorrect. Any suggestions?

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Thank You for your time. I greatly appreciate any suggestions, comments, or criticism. You advice will help me further my knowledge in math. Take care and have a wonderful day.

Answer 1

Suppose there exists some non-zero vector $x_i \in V_i \cap \sum_{j\neq i}V_j$. Then we have $$x_i = \sum_{j \neq i}x_j,$$ for some vectors $x_j \in V_j$, hence $$x_i - \sum_{j \neq i}x_j = 0,$$ Now since $x \neq 0$, the $x_j$ cannot be all zero. This contradicts that $$\underbrace{0 + \cdots + 0}_{\text{$n$ times}} = 0,$$ is the unique decomposition of the zero vector.

Answer 2

I think you yourself gave the proof of $2 \rightarrow 3$:

Take any $\alpha \in V_i\cap\sum_{j\neq i}V_i$.

Then $\alpha \in V_i$ and $\alpha = \alpha_1 + ... + \alpha_{i-1} + \alpha_{i+1} + ... + \alpha_{n}$ for $\alpha_j \in V_j$.

I'm not sure where you got stuck in your argument, now you just can apply your assumption $2$:

$0 = \alpha_1 + ... + \alpha_{i-1} - \alpha + \alpha_{i+1} + ... + \alpha_{n}$ gives a linear combination of $0$ and hence all $\alpha_j$ and $\alpha$ are $0$.