Proof that the $\dim (U+W)= \dim U + \dim W -\dim (U\cap W)$

Question

The Book is Bosch Linear Algebra.

I have difficulties to understand how one deduces this statement from prior statements proved earlier:

Statement already proved:

Definition 1.

Let $U_1,....,U_r$ be linear subspaces of a $KVR$ $V$. The sum is defined as

$$\sum_{i=1}^{r}U_i=\{\sum_{i=1}^{r}b_i;b_i\in U_i, i=1,...,r \}$$

Theorem 2Let $U=\sum_{i=1}^{r}U_i$ be a sum of subspaces of a $KVR$ $V$, the following statements are equivalent:

(i) If $b$ is written as $b=\sum_{i=1}^{r}$ then the $b_i$ are unique. (ii) If $\sum_{i=1}^{r}b_i=0$ then $b_i=0$ for all $i$ (iii) $U_p\cap\sum_{i\neq p}U_i=0$ for all $p=1,...,r$.

Definition 3 If the sum $U=\sum U_i$ fullfills one of the conditions above in Theorem 2 the sum is called direct.

Theorem 4Let $V$ be a $KVR$ $V$ and $U$ a subspace then there exissts a subspace $U'$ with $V=U+U'$ such that the sum is direct and for all such $U'$ the dimension formula:

$$\dim V= \dim U +\dim U'$$ holds

Theorem 5 $\dim (U+W)= \dim U + \dim W -\dim (U\cap W)$

There is a question about the proof, up to a certain point I understand it:

We suppose that $U$ and $U'$ are of finite dimension subspaces of a vectorspace in $V$.

$U+U'$ thus has a finite generating system and thus a finite basis. One chooses in accordance to Theorem 4 a complement $W$ of $U\cap U'$ and $W'$ of $U\cap U'$ such that

$U=(U\cap U') + W$ where the sum is direct and $U'=(U \cap U') + W'$ where the sum is also direct.

Then

$U+U'=(U\cap U')+W+W'$ and the sum (on the RHS) is also direct.

Let $a+b+b'=0$ $a\in$ intersection $b\in W$ and $b'\in W'$ then

$b=-(a+b')\in (U\cap U')+W'=U'$ and because of $b\in W\subset U$. $b\in (U\cap U')\cap W=0$

analogously $b'=0$ then also $a=0$. Therefore the sum is direct

And because of Theorem 4 we now can say

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

This is a sum with $3$ summands, if it would be a sum with two summands then I would understand it. But when it is three sumands I don't know how you can directly deduce from Theorem 4 this dimension formula. You can prove it by induction but the author says this can be directly deduced from theorem $4$ and didn't even mention induction.

I know there are many questions about this formula but I want to understand this specific proof of this formula.

Answer 1

You can use Theorem 4 twice: Let $$A:=W \oplus W'.$$

Note that $A$ is also a vector space and that the sum is direct. So $\dim A = \dim W + \dim W'$ by Theorem 4.

Using Theorem 4 again, we see that $\dim (U+U')=\dim A +\dim(U\cap U')=\dim(U\cap U') +\dim W + \dim W'$.

Answer 2

There is a proposition which states that if $A,B,C$ are subspaces of $V$, then $V=A\oplus B \oplus C \iff V=A+B+C \text{ and } A\cap(B+C)= B\cap(A+C)= C\cap(A+B)=\{0_V\}$

This means that if $V_1=A+B$ and $V_2=C$, then $V=V_1\oplus V_2$ and $\text{dim}V=\text{dim}V_1+\text{dim}V_2$. Moreover, if $$V_1=A\oplus B,$$ then $\text{dim}V_1=\text{dim}A+\text{dim}B$, hence \begin{equation}\tag{1} \text{dim}V=\text{dim}V_1+\text{dim}V_2=\text{dim}A+\text{dim}B+\text{dim}C. \end{equation}

Take now $V=U+U'$, $A=U\cap U'$, $B=W$, $C=W'$ and $V_1=A+B=(U\cap U')+W$. Then $V=U+U'=(U\cap U')\oplus W \oplus W'=A\oplus B \oplus C$. But since you have proved that $V_1=(U\cap U')+W=A+B$ is a direct sum, then $V_1=A\oplus B$, and so by using $(1)$ you have $$\text{dim}(U+U')=\text{dim}V=\text{dim}A+\text{dim}B+\text{dim}C=\text{dim}(U\cap U')+\text{dim}W+\text{dim}W'.$$

Is that what are you asking?