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Linear Algebra: intersection of subspaces 
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#1
Mar712, 01:38 PM

P: 4

1. The problem statement, all variables and given/known data
I'm working on a problem that involves looking at the dimension of the intersection of two subspaces of a vector space. 2. Relevant equations [itex]M \subset V [/itex] [itex]N \subset V [/itex] dim(M [itex]\cap[/itex] N) [itex][\vec{v}]_{B_M}[/itex] is the coordinate representation of a vector v with respect to the basis for M 3. The attempt at a solution I reformulated M [itex]\cap[/itex] N in a bunch of different ways that would be too long to copy down here, but I finally came to this (which may or may not be useful to me in my larger problem but I'm wondering if it is valid itself): [itex]\vec{v}[/itex] is itself, so it must have the same dimension in both M and N, and since the bases are ordered, for each [itex]\vec{b}_{Mi}[/itex] in [itex]B_M[/itex] for which the corresponding scalar is not zero in the linear combination of elements of [itex]B_M[/itex] equal to [itex]\vec{v}[/itex], and each [itex]\vec{b}_{Nj}[/itex] in [itex]B_N[/itex] for which the corresponding scalar is not zero in the linear combination of elements of [itex]B_N[/itex] equal to [itex]\vec{v}[/itex], if i=j then [itex]\vec{b}_{Mi}[/itex] and [itex]\vec{b}_{Nj}[/itex] are dependent and [itex][\vec{v}]_{B_M}[/itex] has zeros in the same places as [itex][\vec{v}]_{B_N}[/itex] but there is a major problem here with the fact that we may have dimM ≠ dimN 


#2
Mar712, 04:07 PM

P: 312

first imaging a basis for the intersection {b1,b2,...,bp}, then expand it to a basis of M, {b1,b2,...,bp,m1,m2,...,mq},
also expand it to a basis for N, {b1,...,bp,n1,n2,...,nr}, dimension of the intersection is p, dim(M)=p+q, dim(N)=p+r. 


#3
Mar812, 12:34 PM

P: 4

beautiful



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