How to prove that the dim of two subspaces added together equals dim

1. Mar 30, 2010

gutnedawg

How to prove that the dim of two subspaces added together equals dim of their union plus 1 iff one space is a subest of the other

In other words,
subspaces: V, S of Vector space: W

$$dim(V+S) = dim(V \cap S) +1$$

if $$V \subseteq S$$ or $$S \subseteq V$$

2. Mar 30, 2010

eok20

Re: Prove

I dont think this is true. If V=S then arent V+S and V intersect S the same?

3. Mar 30, 2010

jbunniii

Re: Prove

eok20 is correct, the statement is false.

What is true is

$$dim(V+S) = dim(V) + dim(S) - dim(V \cap S)$$

which reduces to a simpler form if $V \subseteq S$ or $S \subseteq V$.

4. Mar 30, 2010

gutnedawg

Re: Prove

perhaps I mistyped the question.
$$dim(V+S) = dim(V \cap S) +1$$ is given for these subspaces.

One has to prove that

$$V \subseteq S$$ or $$S \subseteq V$$

5. Mar 30, 2010

jbunniii

Re: Prove

OK, suppose

$$V \not\subseteq S$$ and $$S \not\subseteq V$$.

Then V contains at least one element not in S, hence at least one element not in $V \cap S$. Thus $dim(V) > dim(V \cap S)$, and since dimensions are integers, this is the same as $dim(V) \geq dim(V \cap S) + 1$.

Similarly, S contains at least one element not in V, so $dim(S) \geq dim(V \cap S) + 1$.

Now apply those inequalities to

$$dim(V+S) = dim(V) + dim(S) - dim(V \cap S)$$

6. Mar 30, 2010

gutnedawg

Re: Prove

How do I come to the fact that

$$dim(V+S) = dim(V) + dim(S) - dim( V \cap S)$$

7. Mar 30, 2010

jbunniii

Re: Prove

This is a standard result that should be in just about any linear algebra textbook. E.g., theorem 2.18 in Sheldon Axler's "Linear Algebra Done Right." See page 33 in this online preview: