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Linear algebra - dimension and intersection

  • Thread starter zwingtip
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  • #1
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Homework Statement


let [tex]V[/tex] be a finite dimensional vector space of dimension n. For [tex]W \leq V [/tex] define the codimension of [tex]W[/tex] in [tex]V[/tex] to be [tex]codim(W) = dim(V) - dim(W)[/tex]. Let [tex]W_i, 1 \leq i \leq r[/tex] be subspaces of [tex]V[/tex] and [tex]S = \cap_{i=1}^{r}W_i[/tex]. Prove:

[tex]codim(S) \leq \sum_{i=1}^{r} codim(W_i)[/tex]


Homework Equations


[tex] dim(U + V) = dim(U) + dim(V) - dim(U \cap V)
\sum_{i=1}^{r} codim(W_i) = \sum_{i=1}^{r} (n - dim(W_i)[/tex]

The Attempt at a Solution



I'm completely lost here. I know I need to prove this by induction. Any tips to point me in the right direction? Can I use the fact that [tex]codim(S) = n - dim(\cap_{i=1}^{r} W_i[/tex] and [tex] dim(U \cap V) < dim(U) + dim(V)[/tex] assuming [tex]dim(U + V) \neq 0[/tex]?
 

Answers and Replies

  • #2
lanedance
Homework Helper
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only had a quick look, but not convinced this is true..

take the case when say dim(W1) = dim(W2) = 1, in Rn, then codim(W1) = codim(W2) = n-1

now say the intersection S is only the zero vector then dim(S) = 0 and codim(S) = n >n-1> Can up with oher higherdimensional arguments similarly
 
  • #3
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Still doesn't make sense to me. What if the intersection contains more than just the zero vector? Sorry for being stupid. Help?
 
  • #4
lanedance
Homework Helper
3,304
2
actually i think my exampel was wrong as
codim(S) = n < codim(W1) + codim(W1) = 2n-2, which works for for n>=3

now take R3, say you have 2 planes P1 & P2 that intersect in a line L, then
dim(R3) = 3
dim(P1) = dim(P2) = 3, then codim(P1) = codim(P2) = 2
dim(L) = 1, codim(L) = 2

so
codim(L) = 2 <= codim(P1) + codim(P2) = 1 + 1 = 2
 
  • #5
20
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I think I've got it with an induction proof from the base cases r=1, r=2 and the inductive step being that it holds for r>2 then manipulating it algebraically for an entire page of 5mm graph paper. THanks.
 

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