Proving T is a Scalar Multiple of the Identity Operator: Invariant Subspaces and Linear Operators

Introduction

The study of linear operators in vector spaces is fundamental in linear algebra and functional analysis. A common question in this field is to understand under what conditions a linear operator can be expressed as a scalar multiple of the identity operator. In this article, we explore a specific case where every subspace of V with dimension dim V - 1 is invariant under the linear operator T. We will provide a detailed proof that T is indeed a scalar multiple of the identity operator.

Understanding the Problem

Consider the finite-dimensional vector space (V) over a field (F) with dimension (n text{dim } V). We are given that every subspace (W subseteq V) with dimension (n - 1) is invariant under (T), meaning that (TW subseteq W). Our goal is to prove that the linear operator (T) is a scalar multiple of the identity operator (I).

Step-by-Step Proof

Step 1: Implications of Invariance

Let's start by understanding the implications of the invariance property. Given that (W subseteq V) with (dim W n - 1) is invariant under (T), it means that for any vector (w in W), the image (Tw) is also in (W).

Step 2: Choosing an Appropriate Basis

To proceed with the proof, we select a basis ({ e_1, e_2, ldots, e_n }) for (V). We will analyze the action of (T) on each basis vector. Since (W text{span}{e_1}) is one-dimensional and not our primary focus, we consider the larger space (V setminus text{span}{e_1}) which has dimension (n - 1).

Step 3: Considering a One-Dimensional Subspace

Consider the one-dimensional subspace (W text{span}{e_1}). Although (dim W 1), it is useful to think about (V setminus W) which has dimension (n - 1). The action of (T) on the whole space will respect the structure given by the invariance of (n - 1)-dimensional subspaces.

Step 4: Constructing Invariant Subspaces

Next, we construct an invariant subspace (U text{span}{e_2, e_3, ldots, e_n}) of dimension (n - 1). By the assumption, (TU subseteq U). This means that (T) can be expressed in terms of the coordinates of the vectors in (U).

Step 5: Expressing (Te_1)

Let's express (Te_1) in terms of the basis vectors. Denote (Te_1 a_1 e_1 b_2 e_2 b_3 e_3 ldots b_n e_n). Since (T) maps vectors in (U) back to (U), we must have (b_2 b_3 ldots b_n 0). Otherwise, if any (b_i eq 0), then (Te_1) would not lie in (U) contradicting the invariance property.

Step 6: Concluding for All Basis Vectors

From the above reasoning, we have (Te_1 a_1 e_1). By repeating this process for (e_2, e_3, ldots, e_n), we can consider the subspace (U_i text{span}{e_1, ldots, e_{i-1}, e_{i 1}, ldots, e_n}) which is invariant under (T). This implies:

[Te_i a_i e_i quad text{for some } a_i in F.]

Step 7: Showing All (a_i) Are Equal

Since the choice of (i) was arbitrary, we can assume without loss of generality that for any two basis vectors (e_i) and (e_j), the linearity of (T) implies that (T) acts uniformly across all basis elements. Therefore, (a_i a_j) for all (i, j).

Let (c a_1 a_2 ldots a_n). Hence, we conclude:

[Tv c v quad text{for all } v in V.]

which means that (T) is a scalar multiple of the identity operator, denoted as (T cI).

Conclusion

In conclusion, we have shown that if every subspace of (V) with dimension (dim V - 1) is invariant under (T), then (T) must indeed be a scalar multiple of the identity operator. This result provides deep insights into the structure of linear operators and the significance of invariant subspaces in linear algebra.