Proof dim(U+V)=dim U+dim V - dim(U∩V)

In summary: There are three in U, four in V, and one in U\cap V. Therefore, the dimension of U\cap V is three plus the dimension of V, or 5.
  • #1
b00tofuu
11
0
let U and V be subspaces of Rn. Prove that dim(U+V)=dim U+dim V - dim(U∩V)
 
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  • #2
Okay, first, what is your definition of "dimension" of a vector space?
 
  • #3
its the number of vector in any basis for a subspace
 
  • #4
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?
 
  • #5
a point...?
but i still don't understand how to write the proof... T_T
 
  • #6
the zero point
 
  • #7
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.
 
  • #8
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.
 
  • #9
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.
 
  • #10
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 :-)
 
  • #11
Yes, but for proving that identity, not all all basis will do. Start with [tex]U \cap V[/tex].
 
  • #12
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?
 

FAQ: Proof dim(U+V)=dim U+dim V - dim(U∩V)

What is the meaning of "Proof dim(U+V)=dim U+dim V - dim(U∩V)"?

This equation is known as the "dimension formula" and it helps us understand the relationship between the dimensions of two subspaces and their intersection. It states that the dimension of the sum of two subspaces is equal to the sum of their individual dimensions minus the dimension of their intersection.

How is this equation used in linear algebra?

This equation is used to simplify calculations involving the dimensions of subspaces. It also helps us understand the properties of subspaces and how they relate to each other.

Can you provide an example of how this equation is applied?

Sure, let's say we have two subspaces U and V in a vector space of dimension 3. If the dimension of U is 2 and the dimension of V is 1, then according to the equation, the dimension of U+V (the sum of U and V) would be 2+1-1=2. This means that U+V is a subspace of dimension 2, and we can use this information to make further calculations or conclusions about U and V.

What does the "U∩V" in the equation represent?

The "U∩V" represents the intersection of the two subspaces U and V. This refers to the set of all vectors that are contained in both U and V. In other words, it is the common elements between the two subspaces.

Is this equation always true for any two subspaces?

No, this equation is only true for finite-dimensional subspaces. In infinite-dimensional spaces, the equation does not hold. Additionally, the equation may not hold if one of the subspaces is not a true subspace (e.g. if it does not contain the zero vector).

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