Suppose that

M={(a b ;a,b,c belong to R

-b c)

and N={(x 0 ;x,y belong to R

y 0)

are subspaces of M2R. Show that (M:R) + (N:R) = ((M'intersect'N):R) + ((M+N):R)

Cheers, Shaun

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- Nov 27th 2006, 05:27 AMShaun Gillhelp needed with a proof
Suppose that

M={(a b ;a,b,c belong to R

-b c)

and N={(x 0 ;x,y belong to R

y 0)

are subspaces of M2R. Show that (M:R) + (N:R) = ((M'intersect'N):R) + ((M+N):R)

Cheers, Shaun - Nov 27th 2006, 08:47 AMThePerfectHacker
I presume,

$\displaystyle M+N=M\cup B$

This looks like*the principle of inclusion-exlusion*.

Thus,

$\displaystyle |M + N|=|M|+|N|-|M\cap N|$

Thus,

$\displaystyle |M + N|+|M\cap N|=|M|+|N|$

So dimensions are,

$\displaystyle (M+N:R)+(M\cap N:R)=(M:R)+(N:R)$

I believe that is the idea.

But the sad thing is I do not know how to generalize this to infinite fields. So far this confirms what you said for finite fields. - Nov 27th 2006, 09:00 AMPlato
I think that these are the same problem.

http://www.mathhelpforum.com/math-he...ons-basis.html