Linear Algebra: intersection of subspaces

TheTangent
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Homework Statement


I'm working on a problem that involves looking at the dimension of the intersection of two subspaces of a vector space.


Homework Equations


M \subset V
N \subset V
dim(M \cap N)
[\vec{v}]_{B_M} is the coordinate representation of a vector v with respect to the basis for M



The Attempt at a Solution


I reformulated M \cap 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):

\vec{v} is itself, so it must have the same dimension in both M and N, and since the bases are ordered, for each \vec{b}_{Mi} in B_M for which the corresponding scalar is not zero in the linear combination of elements of B_M equal to \vec{v}, and each \vec{b}_{Nj} in B_N for which the corresponding scalar is not zero in the linear combination of elements of B_N equal to \vec{v}, if i=j then \vec{b}_{Mi} and \vec{b}_{Nj} are dependent

and [\vec{v}]_{B_M} has zeros in the same places as [\vec{v}]_{B_N}

but there is a major problem here with the fact that we may have dimM ≠ dimN
 
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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.
 
beautiful
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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