Proving T is Scalar Multiple of Identity Operator: Invariant Subspace

In summary, the conversation discusses how to prove that a linear operator on a finite dimensional vector space is a scalar multiple of the identity operator if every subspace of dimension dim V-1 is invariant under it. The conversation suggests starting by considering subspaces U and W with the same dimension, and showing that their intersection is T-invariant. Then, by fixing a basis and considering how the operator acts on certain vectors, it can be shown that all eigenvalues of T are equal, thus proving that T is a scalar multiple of the identity operator.
  • #1
cubixguy77
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Homework Statement


Suppose T is a linear operator on a finite dimensional vector space V, such that every subspace of V with dimension dim V-1 is invariant under T. Prove that T is a scalar multiple of the identity operator.

The Attempt at a Solution


I'm thinking of starting by letting U and W be subspaces of V with dim U = dim W = dim V-1
This means that U and W are invariant under T. Good start? Where do i go from there to show that T is a scalar multiple of the identity operator?
 
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  • #2
Let n = dim V, v some element of V, <v> the 1-dim subspace generated by v.

Show that there are n-1 subspaces of dimension n-1 such that their intersection is <v>. Conclude that <v> is T-invariant.
Now you can fix a basis [itex]\{e_i\}[/itex]. From what you have shown you know [itex]T e_i =\lambda_i e_i[/itex].

You want to show that all [itex]\lambda_i[/itex] are equal. For this, consider for example [itex]e_i+e_{i+1}[/itex]. Since T fixes evey one-dimensional subspace it has to act on this vector as scalar multiplication, [itex]T(e_i+e_{i+1})=\nu_i^{i+1}(e_i+e_{i+1})[/itex].By linearity you also know that [itex]T(e_i+e_{i+1})=\lambda_i e_i+\lambda_{i+1}e_{i+1}[/itex]. Using the linear independece of the chosen basis you can conclude [itex]\lambda_i=\nu_i^{i+1}=\lambda_{i+1}[/itex]
 

1. What is an invariant subspace?

An invariant subspace is a vector subspace of a given vector space that is invariant under a given linear transformation. This means that any vector within the subspace, when transformed by the linear operator, will remain within the subspace.

2. How is the identity operator related to invariant subspaces?

The identity operator is a linear transformation that leaves all vectors unchanged. This means that any vector within an invariant subspace will remain unchanged when transformed by the identity operator.

3. How can we prove that T is a scalar multiple of the identity operator for an invariant subspace?

To prove that T is a scalar multiple of the identity operator for an invariant subspace, we need to show that T multiplied by a scalar value will give the same result as the identity operator. This can be done by applying T to a vector within the subspace and comparing it to the result of multiplying the same vector by the scalar value and then applying the identity operator.

4. What is the significance of proving that T is a scalar multiple of the identity operator for an invariant subspace?

Proving that T is a scalar multiple of the identity operator for an invariant subspace is important because it shows that the linear transformation T has a special property - it leaves all vectors within the subspace unchanged. This can make certain calculations and proofs easier to perform.

5. Is it possible for a linear transformation to have more than one invariant subspace?

Yes, it is possible for a linear transformation to have more than one invariant subspace. In fact, every linear transformation has at least two invariant subspaces - the entire vector space and the zero subspace. However, some linear transformations may have additional invariant subspaces depending on the specific properties of the transformation.

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