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May 11th, 2018, 01:30 AM  #1 
Newbie Joined: Mar 2018 From: Split, Croatia Posts: 13 Thanks: 0  Sum of two invariant subspaces
Let f : V>V be linear mapping and L,M finvariant subspaces of V. Prove that L+M is also finvariant. I know how things work for special case when L is the direct complement of M, but I'm stuck with this L+M. 
May 11th, 2018, 07:16 AM  #2 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
Suppose that T:V→V is a linear transformation with eigenvalue λ and associated eigenspace ET(λ). Let W be any subspace of ET(λ). Then W is an invariant subspace of V relative to T. http://linear.ups.edu/version3/scla/sectionIS.html 
May 11th, 2018, 10:02 PM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 598 Thanks: 366 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
The exercise is essentially one line. Can you explain how you are showing this for the case that $L \oplus M = V$? I think you should find that you never need that assumption.  
May 13th, 2018, 06:19 AM  #4 
Newbie Joined: Mar 2018 From: Split, Croatia Posts: 13 Thanks: 0  I think I got the idea
I finally proved this, hope it is ok? Thank you

May 13th, 2018, 08:47 AM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 598 Thanks: 366 Math Focus: Dynamical systems, analytic function theory, numerics 
yep you got it.

May 14th, 2018, 08:27 AM  #6  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Quote:
Definition: A subspace W of a vector space V is said to be invariant with respect to a linear transformation T ∈ L(V, V ) if T (W) ⊆ W. * The definition virtually implies the answer: If x ∈ U and y ∈ W, T(ax+by)=aTx+bTy belongs to U+W because aTx belongs to U and bTy belongs to W, by definition. And U+W is a subspace, because if ax1=ay1+az1 and bx2 =by2+bz2 belong to U+W, so does ax1+bx2=a(y1+y2)+b(z1+z2). * https://math.okstate.edu/people/bine...35023l18.pdf  

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