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May 11th, 2018, 02:30 AM   #1
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Sum of two invariant subspaces

Let f : V->V be linear mapping and L,M f-invariant subspaces of V. Prove that L+M is also f-invariant.
I know how things work for special case when L is the direct complement of M, but I'm stuck with this L+M.
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May 11th, 2018, 08:16 AM   #2
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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/section-IS.html
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May 11th, 2018, 11:02 PM   #3
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Quote:
Originally Posted by zylo View Post
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/section-IS.html
Not every invariant subspace is spanned by eigenvectors.

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.
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May 13th, 2018, 07:19 AM   #4
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I think I got the idea

I finally proved this, hope it is ok? Thank you
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May 13th, 2018, 09:47 AM   #5
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yep you got it.
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May 14th, 2018, 09:27 AM   #6
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Quote:
Originally Posted by Birgitta View Post
Let f : V->V be linear mapping and L,M f-invariant subspaces of V. Prove that L+M is also f-invariant.
I acknowledge Brigitta's correct answer. However, I think it should start with the definition of invariant subspace.

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...3-5023-l18.pdf
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