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
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.
I think that these are the same problem.
http://www.mathhelpforum.com/math-he...ons-basis.html