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Is there a way to generalize $\dim(\sum_{i=1}^{k}W_i)$ from the following formula?

$\dim(U+W)=\dim U+\dim W-\dim (U \cap W)$

I tried looking at examples

$k=3$:

$\dim(W_1+W_2+W_3)=\dim(W_1+W_2)+\dim W_3-\dim((W_1+W_2) \cap W_3)=$ $\dim W_1+\dim W_2+\dim W_3-\dim(W_1\cap W_2)-\dim((W_1+W_2) \cap W_3)$

$k=4$:

$\dim(W_1+W_2+W_3+W_4)=$

$\dim (W_1+W_2+W_3)+\dim W_4-\dim ((W_1+W_2+W_3)\cap W_4)$=

$\dim W_1+\dim W_2+\dim W_3+\dim W_4-\dim(W_1\cap W_2)+ \cdots$

$\cdots -\dim((W_1+W_2) \cap W_3)-\dim ((W_1+W_2+W_3)\cap W_4)$

So I guess $\dim(\sum_{i=1}^{k}W_i)=\sum_{i=1}^{k}\dim(W_i)-\sum_{i=1}^{k}\dim((\sum_{j=1}^{i-1}W_j)\cap W_i)$. Is it true? If so, how do I prove it?

Also if it is true, can I conclude $\dim(\bigoplus_{i=1}^k W_i) =\sum_{i=1}^{k}\dim W_i$? How?

Theorem
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1 Answers1

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Thanks for @daw for pointing out my error in my previous answer. For general case $n$, note that $W_2+\dots +W_n$ is a subspace, thus using the formula for 2 dimensions gives:

$$\dim \left(\sum_{j=1}^{n}W_j\right) =\dim(W_1)+\dim(W_2+\dots + W_n) -\dim(W_1 \cap \sum_{k=2}^{n}W_k)$$

So expanding all the way down by repeatedly using the formula gives:

$$\dim \left(\sum_{j=1}^n W_j \right)=\sum_{k=1}^n \dim(W_k) -\sum_{m=1}^{n-1}\dim\left(W_m \cap \sum_{i=m+1}^{n}W_i\right)$$

Now to prove the second statement, we want to show that for any integer $m<n$, the following is true:

$$W_m \cap \sum_{i=m+1}^n W_i=\{\vec 0\}$$

First notice that if $\bigoplus_{k=1}^n W_k$ makes sense, then $\bigoplus_{k=m}^n W_k$ must make sense too.

Then we will use the following lemma: if $U\bigoplus V$ makes sense, then $U\cap V=\{0\}$

Proof: Since any vector in $U\bigoplus V$ is uniquely represented by sum of vector $\vec u\in U$ and $\vec v\in V$. Consider $\vec 0=\vec u +(-\vec u)$, it implies that if some $\vec u\in U$, then $-\vec u\in V$, by definition of vector space we know that $u \in V$. But $\vec u= -\vec u=\vec 0$ is the unique solution. Thus we have $U\cap V=\{0\}$ precisely.

Applying the lemma, we have $$W_m\cap \left(\bigoplus_{k=m+1}^n W_k\right)=\{0\}$$

because $$\bigoplus_{k=m+1}^n W_k =\sum_{k=m+1}^n W_k$$,

for each $m$. Completing the proof.

Note: you can prove the second statement with a simpler proof involving finding bases for each $W_j$ and put them in a list.

lEm
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