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

[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 $$S \subseteq V$$

 

[eok20]

I don't think this is true. If V=S then arent V+S and V intersect S the same?

 

[jbunniii]

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 $S \subseteq V$.

 

[gutnedawg]

I don't think this is true. If V=S then arent V+S and V intersect S the same?

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 $$S \subseteq V$$

 

[jbunniii]

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 $$S \subseteq V$$

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

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

to achieve a contradiction.

 

[gutnedawg]

How do I come to the fact that

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

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

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

to achieve a contradiction.

 

[jbunniii]

How do I come to the fact that

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

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:

http://books.google.com/books?id=BN...esnum=1&ved=0CDoQ6AEwAA#v=onepage&q=&f=false"

Here is another proof (PDF file), but it looks more longwinded than it needs to be:

http://www.its.caltech.edu/~clyons/ma1b/intsumdimthm.pdf

[Edit]: See also Problem 16, parts 2 and 3 here:

http://en.wikibooks.org/wiki/Linear_Algebra/Combining_Subspaces