Suppose we have a vector space $V$, and $U$, $W$ subspaces of $V$.

Dimension theorem states: $$ \dim(U+W)=\dim U+ \dim W - \dim (U\cap W).$$

My question is:

Why is $U \cap W$ necessary in this theorem?

  • 2
    $\begingroup$ intersection of U and W. In other words, I think it's the set of vectors which are in both U and W. We don't want to include that when we go about the dimension. $\endgroup$ – Cecile Feb 28 '13 at 22:49
  • $\begingroup$ It's the intersection of the two subspaces. $\mathbb{U}\cap \mathbb{W}=\{v\in \mathbb{V}| v\in\mathbb{W}\text{ and }v\in\mathbb{U}\}$. $\endgroup$ – MichalisN Feb 28 '13 at 22:50

You can think in $A=\{a_1,...,a_n\}$ as base of $U$ and $B=\{b_1,...,b_m\}$ as a base of $V$, we know that the dimension of a vectorial space is the number of elements of some base. If $b_1$ it isn't linear combination of the elements of $A$ so $C_1=A \cup \{b_1$}. If $b_p$ it isn't linear combination of the elements of $C_{p-1}$, so $C_p=C_{p-1}\cup\{b_p\}$. Let $\ D=B -C_{p-1}$, now it is easy to see that $C_m$ is base of $\ U+V$ and $D$ is base of $\ U\cap V$.

This is an algorithm to obtain $U+V$ and $U\cap V$.

Without lost of generality, if $C_m=\{a_1,a_2,...,a_n,b_1,b_2,...,b_{m-k}\} \Rightarrow dim(U+V)=n+k$

on the other hand $dim(U)+dim(V)-dim(U\cap V)=n+m-(m-k)=n+k$


$U \cap W$ is the intersection of the vector spaces $U$ and $W$, that is, the set of all vectors of the space $V$ which are in both subspaces $U$ and $W$.

As $U$ and $W$ are both subspaces of $V$, their intersection $U \cap W$ is also a subspace of $V$ (this assertion can be easily proved). Because $U \cap W$ is a subspace, it is also a vector space itself, and as such it has a basis. The number of elements in this basis will be the space's dimension, $\dim (U \cap W)$.

Loosely speaking, one could think that summing $\dim(U)$ and $\dim(W)$ would yield $\dim(U+W)$. But as $(U \cap W) \subset U$ and $(U \cap W) \subset W$, the sum $\dim(U) + \dim(W)$ "counts" two times the dimension of $U \cap W$ - once in $\dim(U)$ and once more in $\dim(W)$. To make it sum up to $\dim(U+W)$ accurately, we must then subtract the dimension of $U \cap W$, so that it is "counted" only once. This way, we obtain:

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

Note that this is not, by any means, a formal proof. It is only an informal explanation of why $U \cap W$ is needed in this formula.

  • $\begingroup$ I think the question's edition by @Ludolila changed its meaning. Looking at the original version, I had the impression the author wanted to know why U∩W was necessary, not what was the definition of U∩W. I have now added my own edition trying to restore its original meaning. $\endgroup$ – Wheepy Mar 1 '13 at 1:27

It is the intersection of the subspaces: $$U\cap W = \{v\in V | v \in U \wedge v\in W \} .$$

  • $\begingroup$ Is that clearly all the deal? I would wait for more answers. $\endgroup$ – Billie Feb 28 '13 at 23:20

Maybe you are looking for a more intuitive answer.

The sum $U+W$ of subspaces $U$ and $W$ is the smallest subspace of $V$ that contains $U$ and $W$. If $U$ and $W$ are not so independent then $U+W$ is very close to both subspaces. This happens just when the intersection $U\cap W$ is big. The extreme case is when $U=W$. In this case $U+W,\ U\cap W,\ U$ and $W$ are all equal and we have: $$dim(U+W)=dim(U)=dim(W)=dim(U\cap W)$$ On the other hand, in this case: $$dim(U)+dim(W)=2dim(U+W)$$

I hope this helps you to understand why in the Dimension Theorem to obtain $dim(U+W)$ we have to substract $dim(U\cap W)$ to the sum of the dimensions of both spaces.


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