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Linear Algebra: Direct sum proof 
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#1
May3110, 04:43 PM

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Let U and V be subspaces of a vector space W. If W=U [tex]\oplus[/tex] V, show U [tex]\bigcap[/tex] V={0}.
I'm a bit lost on this one... as I thought this was essentially the definition of direct sum. I'm unsure where to start. Any help would be great! 


#2
May3110, 06:16 PM

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PF Gold
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What's the definition of a direct sum you're using?



#3
May3110, 07:52 PM

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If U and V are subspaces of vector space W, and each w in W can be written uniquely as a sum u+v where u is in U and v is in V then W is a direct sum of U and V. 


#4
May3110, 08:00 PM

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Linear Algebra: Direct sum proof
So were you able to do the problem?



#5
May3110, 08:04 PM

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#6
May3110, 08:12 PM

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Well, surely you can do something on it  even if it's just rewriting the problem in a less opaque form.
e.g. do you know anything about proving two subspaces equal? (Or two sets?) 


#7
Jun110, 06:03 AM

P: 317

If U and V are subspaces of a Vector space W and each [tex]w \in W[/tex] can be written as the unique sum as u+v where [tex]u \in U[/tex] and [tex]v \in V[/tex] then W is the direct sum of U and V and can be written [tex]W = U \oplus V[/tex] 


#8
Jun110, 06:41 AM

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Suppose there were a nonzero vector, w, in both U and V and let u be any vector in U. Now, write u as two different sums of vectors in U and V.



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