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gravenewworld
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Does every linear operator have a nontrivial invariant subspace? My professor mentioned this question in class, but never actually answered it. I am curious if this is true or not and why.
Sounds like first year algebra. One of the most important things you'll be dealing with in this algebra course, and probably the next (and you'll see it in a number of other places as well) are eigenvalues and eigenvectors. Given some operator T, we call [itex]\lambda[/itex] and eigenvalue of T if there exists some vector [itex]v \in V[/itex] such that [itex]T(v) = \lambda v[/itex], and [itex]v[/itex] is called the eigenvector of T corresponding to [itex]\lambda[/itex]. If T has an eigenvalue, then it has an eigenvector, and if we let that vector be v, then clearly Span{v} is a T-invariant subspace (you can easily check this for yourself).gravenewworld said:Okay I will buy that if you work with R^n, but what if you start going into the complex number land as your vector space? Then T(a, ia)=-(ia,a)=-i(a,ia). I reread my notes and my professor changed the original question to an the equivalent question of Given T an element of L(V); where dim(V)>=1 Does there exist a nonzero vector v and element of V such that T(v)=(lambda)v for lambda an element of the field which the vector space is over. If you use C then the example you gave works.
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A linear operator is a mathematical function that operates on vector spaces, transforming one vector into another through specific rules of addition and scalar multiplication.
An invariant subspace is a subset of a vector space that remains unchanged under the action of a given linear operator. In other words, the vectors within this subspace remain within the subspace after being operated on by the linear operator.
No, not every linear operator has a nontrivial invariant subspace. Some linear operators may only have the trivial invariant subspace, which consists of the zero vector.
To determine if a linear operator has a nontrivial invariant subspace, you can use the Cayley-Hamilton theorem. If the characteristic polynomial of the linear operator has a root of multiplicity greater than 1, then there exists a nontrivial invariant subspace.
The existence of a nontrivial invariant subspace is important in linear algebra because it allows for the simplification of certain computations and theorems. It also helps in understanding the behavior of a linear operator and its relationship with the underlying vector space.