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I'm quite confused about the direct sum.

Let $V$ be a finite dimensional vector space and $W_i$ its subspaces.

Firstly can anyone give a counterexample of

$dim(V)=dim(W_1)+dim(W_2)+...+dim(W_k) \implies V=W_1\oplus W_2\oplus...\oplus W_k$ ?

I know that the opposite implication is true of course, but I cannot understand how this one must be false.

Secondly I don't understand why if I add the condition $V=W_1+W_2+...+W_k$

then this becomes true, infact I know that

$V=W_1\oplus W_2\oplus...\oplus W_k \iff V=W_1+W_2+...+W_k \wedge dim(V)=dim(W_1)+dim(W_2)+...+dim(W_k)$

In general isn't the condition $V=W_1+W_2+...+W_k$ automatically implied by $dim(V)=dim(W_1)+dim(W_2)+...+dim(W_k)$ ?

I mean, supposing $dim(V)=dim(W_1)+dim(W_2)+...+dim(W_k)$ but $V\neq W_1+W_2+...+W_k$ then there would be at least a vector $v$ in $V$ but not in $W_1+W_2+...+W_k$ that spans a one dimensional vector space $\mathscr{L}(v)$, which is a contradiction because $dim(V)=dim(W_1)+dim(W_2)+...+dim(W_k)$

Am I missing something?

Thanks a lot for your help

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Let me give you a simple example. Suppose we have a $3$-dimensional space $V$, and $L$ is a $1$-dimensional subspace. Set $L_1=L_2=L_3=L$. Then the dimensions add up, yet $V$ is not the direct sum of $L_1,L_2,L_3$. Technically it is isomorphic to this direct sum as a vector space, but that's not the same as actually being the (internal) direct sum.

The implication fails any time two of the subspaces intersect nontrivially; then the sum simply can't be direct.

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    $\begingroup$ It fails more generally if any subspace $W_i$ has nontrivial intersection with the span of some other $W_{j_1},W_{j_2},\ldots,W_{j_k}$. $\endgroup$ – Matt Samuel Dec 13 '15 at 1:04
  • $\begingroup$ Thanks for the answer! So the condition $V=W_1+W_2+...+W_k$ of course does not guarantee that any intersection is trivial but togheter with the fact that $dim(V)=dim(W_1)+dim(W_2)+...+dim(W_k)$ does it imply that $V=W_1\oplus W_2\oplus...\oplus W_k$? $\endgroup$ – Gianolepo Dec 13 '15 at 9:08
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    $\begingroup$ Yes it does when the dimensions add up. $\endgroup$ – Matt Samuel Dec 13 '15 at 15:19
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You forget that we only know $\displaystyle\dim\Bigl(\sum_i W_i\Bigr)\leq\sum_i\dim W_i$. Equality would precisely mean the sum is direct.

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