(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Linear Algebra: intersection of subspaces

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