Proof dim(U+V)=dim U+dim V - dim(U∩V)

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let U and V be subspaces of Rn. Prove that dim(U+V)=dim U+dim V - dim(U∩V)
 
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its the number of vector in any basis for a subspace
 
Here's a hint: if [tex]U \cap V = \left\{0\right\}[/tex], then what must be the intersection of any base of U with any base of V?
 
Notice that the basis are SETS of vectors: if [tex]U \cap V = \left\{0\right\}[/tex], then the intersection of any base of U, with any base of V will be the empty set; also, [tex]dim\left(U\capV\right) = dim\left(\left\{0\right\}\right) = 0[/tex]. Try this particular case first, then see if can generalize when [tex]U \cap V[/tex] is not the null subspace.
 
One way to approach the problem is to ask yourself if you can find bases for the vector spaces U + V, U, V, and U ∩ V that are related somehow to each other.
 
You are complicating too much; it's simpler than that: consider a basis [tex]\left\{b_{i}\right\}[/tex] for [tex]U \cap V[/tex]; this basis can be extended to a basis [tex]\left\{u_{i}\right\}[/tex] of U and [tex]\left\{v_{i}\right\}[/tex] of V; now it's only a matter of counting the vectors.
 
I believe we're talking about the same thing. Extending a basis is what I meant by finding bases for U, V, and U ∩ V that are related to each other, then picking the right vectors from those bases to be a basis for U + V. Of course, I still might be missing something even simpler; it wouldn't be the first time :-)
 
Yes, but for proving that identity, not all all basis will do. Start with [tex]U \cap V[/tex].
 
Choose a basis for U. If any of those basis vectors are also in V, you can construct a basis for V including those vectors. If not, just choose any basis for V. Of course, the vectors in both bases, if any, form a basis for [itex]U\cap V[/itex]. Now, just count!

How many vectors are there in the basis for U? How many vectors are there in the basis for V? How many are in both?