Question

Let \(T: V \rightarrow V\) be a linear operator where \(V\) is finite dimensional. If \(U \subseteq V\) is a subspace, let \(\bar{U}=\left\\{\mathbf{u}_{0}+T\left(\mathbf{u}_{1}\right)+T^{2}\left(\mathbf{u}_{2}\right)+\cdots+T^{k}\left(\mathbf{u}_{k}\right) \mid \mathbf{u}_{i}\right.\) in \(U, k \geq\) 0\\}. Show that \(\bar{U}\) is the smallest \(T\) -invariant subspace containing \(U\) (that is, it is \(T\) -invariant, contains \(U\), and is contained in every such subspace).

Short answer

\(\bar{U}\) is the smallest \(T\)-invariant subspace containing \(U\).

Step-by-step solution

Understanding T-Invariance

A subspace \(W\) is \(T\)-invariant if for every vector \(\mathbf{w} \in W\), the vector \(T(\mathbf{w})\) is also in \(W\). To prove \(\bar{U}\) is \(T\)-invariant, we need to show that applying \(T\) to any vector in \(\bar{U}\) results in another vector that still belongs to \(\bar{U}\).

Showing \(\bar{U}\) is T-Invariant

Take any vector \(\mathbf{v} = \mathbf{u}_{0}+T(\mathbf{u}_{1})+T^{2}(\mathbf{u}_{2})+\cdots+T^{k}(\mathbf{u}_{k})\) in \(\bar{U}\). Then \(T(\mathbf{v}) = T(\mathbf{u}_{0}) + T^{2}(\mathbf{u}_{1}) + \cdots + T^{k+1}(\mathbf{u}_{k})\), which is still in the form of an element of \(\bar{U}\) (where \(T^{k+1}(\mathbf{u}_k) = 0\) if \(U\) is finite-dimensional), thus remaining in \(\bar{U}\). Hence, \(\bar{U}\) is \(T\)-invariant.

Showing \(U \subseteq \bar{U}\)

By definition, \(\bar{U}\) is generated by linear combinations containing terms \( \mathbf{u}_{0}, T(\mathbf{u}_{1}), \ldots \) where each \(\mathbf{u}_i\) is in \(U\). This naturally implies that \(\mathbf{u}_0 \in U\). Hence, every element of \(U\) can be seen as a vector in \(\bar{U}\), demonstrating that \(U \subseteq \bar{U}\).

Minimality with Respect to T-Invariance

Let \(W\) be any \(T\)-invariant subspace containing \(U\). Since \(\bar{U}\) is generated by \(U\) and the action \( T, T^2, \ldots \) on elements of \(U\), and \(W\) contains \(\mathbf{u}_0, T(\mathbf{u}_1), T^2(\mathbf{u}_2), \ldots\), it follows that \(W\) must contain \(\bar{U}\). Thus, \(\bar{U}\) is contained in any \(T\)-invariant subspace containing \(U\), establishing its minimality.

Key concepts

Linear Transformations

Linear transformations are core elements in the study of linear algebra. They are functions that map from one vector space to another while preserving the operations of vector addition and scalar multiplication. In the context of our exercise, we deal with a linear transformation denoted as \( T: V \rightarrow V \). This means \( T \) takes a vector from the vector space \( V \) and transforms it into another vector within the same space.

Key properties of linear transformations include:

• Additivity: \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \) for any vectors \( \mathbf{u} \) and \( \mathbf{v} \).
• Scalar Multiplication: \( T(c\mathbf{u}) = cT(\mathbf{u}) \) for any vector \( \mathbf{u} \) and scalar \( c \).

In the task, the focus is on determining how these transformations interact with subspaces like \( U \), specifically identifying how transformations can generate larger, invariant subspaces \( \bar{U} \).
Understanding these transformations helps to simplify complex systems by reducing them to operations on smaller subspace components.

Subspace Properties

A subspace is essentially a smaller vector space that exists within a larger one, adhering to specific properties that allow it to retain the structure of a vector space on its own. Subspaces are crucial because they provide a manageable structure to examine within the immense space. The exercise introduces \( U \) as a subspace of the finite-dimensional space \( V \).

Some vital subspace properties include:

• Closure under addition: If \( \mathbf{u} \) and \( \mathbf{v} \) are in the subspace, \( \mathbf{u} + \mathbf{v} \) must also be in the subspace.
• Closure under scalar multiplication: If \( \mathbf{u} \) is in the subspace and \( c \) is a scalar, \( c\mathbf{u} \) must also be in the subspace.
• Contains the zero vector: The zero vector is always part of any subspace.

To understand \( \bar{U} \)'s properties, especially its \( T \)-invariance, it is essential to grasp these basic subspace characteristics. \( \bar{U} \) not only contains \( U \) but is also defined as the smallest \( T \)-invariant subspace enveloping \( U \), making it closed under both vector operations and transformations.

Finite-Dimensional Vector Spaces

Finite-dimensional vector spaces are those that have a finite number of basis vectors, which is a crucial aspect in the study and application of linear algebra. The dimensionality essentially means the number of independent vectors that span the entire space. The concept of a basis is central here.

In this exercise, the vector space \( V \) is finite-dimensional, implying that it can be represented by a finite few vectors that linearly cover the entire space. This characteristic greatly simplifies problems in linear algebra because it ensures the space is not infinite, allowing for complete descriptions and operations like transformation and decomposition.

Studying finite-dimensional vector spaces involves understanding

• Basis and dimension: The smallest set of vectors that span the space gives the dimension of the space.
• Span: All vectors in the space can be expressed as linear combinations of the basis vectors.

In our exercise, because \( V \) is finite-dimensional, operations like \( T^k \) eventually yield results that remain in the span, ensuring that constructs like \( \bar{U} \) are also finite-dimensional. This footing helps find the minimal \( T \)-invariant subspace that contains \( U \).