- #1

- 35

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In other words,

subspaces: V, S of Vector space: W

[tex] dim(V+S) = dim(V \cap S) +1 [/tex]

if [tex]V \subseteq S[/tex] or [tex]S \subseteq V [/tex]

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- Thread starter gutnedawg
- Start date

- #1

- 35

- 0

In other words,

subspaces: V, S of Vector space: W

[tex] dim(V+S) = dim(V \cap S) +1 [/tex]

if [tex]V \subseteq S[/tex] or [tex]S \subseteq V [/tex]

- #2

- 200

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

- #3

- 3,472

- 251

eok20 is correct, the statement is false.

What is true is

[tex]dim(V+S) = dim(V) + dim(S) - dim(V \cap S)[/tex]

which reduces to a simpler form if [itex]V \subseteq S[/itex] or [itex]S \subseteq V[/itex].

- #4

- 35

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

perhaps I mistyped the question.

[tex] dim(V+S) = dim(V \cap S) +1 [/tex] is given for these subspaces.

One has to prove that

[tex] V \subseteq S [/tex] or [tex] S \subseteq V [/tex]

- #5

- 3,472

- 251

perhaps I mistyped the question.

[tex] dim(V+S) = dim(V \cap S) +1 [/tex] is given for these subspaces.

One has to prove that

[tex] V \subseteq S [/tex] or [tex] S \subseteq V [/tex]

OK, suppose

[tex]V \not\subseteq S[/tex] and [tex]S \not\subseteq V[/tex].

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

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

Now apply those inequalities to

[tex]dim(V+S) = dim(V) + dim(S) - dim(V \cap S)[/tex]

to achieve a contradiction.

- #6

- 35

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How do I come to the fact that

[tex] dim(V+S) = dim(V) + dim(S) - dim( V \cap S) [/tex]

OK, suppose

[tex]V \not\subseteq S[/tex] and [tex]S \not\subseteq V[/tex].

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

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

Now apply those inequalities to

[tex]dim(V+S) = dim(V) + dim(S) - dim(V \cap S)[/tex]

to achieve a contradiction.

- #7

- 3,472

- 251

How do I come to the fact that

[tex] dim(V+S) = dim(V) + dim(S) - dim( V \cap S) [/tex]

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 [Broken]

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

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

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