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Let $V$ be a vector space and let $W_1,\ldots,W_k$ be subspaces of $V$ such that $V = W_1 +\cdots+ W_k$ and $\dim(V) = \dim(W_1) +\cdots+ \dim(W_k)$. Prove that $V = W_1 \oplus\cdots\oplus W_k $.

My attempt: $k=1$ is trivial, for $k=2$,

$\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$ which gives us $\dim(W_1 \cap W_2)=0$, which implies $W_1 \cap W_2=\{0\}$. Hence $V = W_1 \oplus W_2 $.

For $k=3$, we get $\dim(W_1+W_2+W_3)=\dim(W_1+W_2)+\dim(W_3)-\dim((W_1+ W_2)\cap W_3)$, which implies $\dim(W_1 \cap W_2)+\dim((W_1+ W_2)\cap W_3)=0$. We get the following:

$1. \,W_1 \cap W_2 = \{ 0 \}$

$2. \, (W_1+ W_2)\cap W_3 = \{0\} $ , $(W_1+ W_3)\cap W_2 = \{0\} $ and $(W_2+ W_3)\cap W_1 = \{0\}$

Are conditions $1$ and $2$ enough to prove that $V = W_1 \oplus W_2 \oplus W_3 $ ? This method isn't good for solving this problem for general $k$, is there any better way to prove this?

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In fact $$\begin{cases} W_1 \cap W_2 = \{ 0 \} \\ (W_1+ W_2)\cap W_3 = \{0\} \end{cases}$$ are sufficient conditions. The second one implies $(W_1+W_2) + W_3= (W_1+W_2) \oplus W_3$ and the first one that $W_1 + W_2 = W_1\oplus W_2$. Therefore $$V= W_1+W_2+W_3 = W_1\oplus W_2 \oplus W_3.$$

However for the general case, I would rather proceed by induction on the dimension $n = \dim V$:

  1. Easy for $n=2$...
  2. Suppose that the result is true for $n \ge 2$ and that $V = W_1 +···+ W_k$ with $\dim V = \displaystyle\sum_{i=1}^{k} \dim W_i = n+1$. Without loss of generality, we can suppose that $\dim W_k \ge 1$. Then $V = (W_1 +···+ W_{k-1})+ W_k$ and $\dim V = \dim (W_1 +···+ W_{k-1})+ \dim W_k$. With what you proved in your question, you have $V = W \oplus W_k$ where $W = W_1 +···+ W_{k-1}$. Now $\dim W \le n$. So you can apply induction hypothesis to it, getting $W = W_1 \oplus···\oplus W_{k-1}$ and finally $V = W_1 \oplus···\oplus W_{k-1} \oplus W_k$ as desired.
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You have a map $T$ from the external direct sum $W_1\oplus\cdots W_k$ to $V$ taking $(w_1,\ldots,w_k)$ to $w_1+\cdots+w_k$. Its image is $W_1+\cdots+W_k=V$. But both $W_1\oplus\cdots W_k$ and $V$ have the same dimension, so $T$ is an isomorphism. The injectivity of $T$ implies that inside $V$, $W_1+\cdots+W_k$ is an internal direct sum.

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