# Linear Algebra: intersection of subspaces

1. Mar 7, 2012

### TheTangent

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

3. 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

2. Mar 7, 2012

### sunjin09

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. Mar 8, 2012

beautiful