Question

Let \(T: V \rightarrow V\) be an operator where \(T^{3}=0 .\) If \(\mathbf{u} \in V\) and \(U=\operatorname{span}\left\\{\mathbf{u}, T(\mathbf{u}), T^{2}(\mathbf{u})\right\\},\) show that \(U\) is \(T\) -invariant and has dimension \(3 .\)

Short answer

\( U \) is \( T \)-invariant, and \( \dim(U) = 3 \).

Step-by-step solution

Define T-invariance

A subspace \( U \) is called \( T \)-invariant if for any vector \( \mathbf{v} \in U \), the vector \( T(\mathbf{v}) \) is also in \( U \).

Check T-invariance for \(U\)

Given that \( U = \text{span}\{ \mathbf{u}, T(\mathbf{u}), T^2(\mathbf{u}) \} \), we need to show that \( T(\mathbf{v}) \in U \) for any vector \( \mathbf{v} = a\mathbf{u} + bT(\mathbf{u}) + cT^2(\mathbf{u}) \). Applying \( T \), we have:\[ T(\mathbf{v}) = T(a\mathbf{u} + bT(\mathbf{u}) + cT^2(\mathbf{u})) = aT(\mathbf{u}) + bT^2(\mathbf{u}) + cT^3(\mathbf{u}) \]Since \( T^3(\mathbf{u}) = 0 \), it simplifies to:\[ T(\mathbf{v}) = aT(\mathbf{u}) + bT^2(\mathbf{u}) \]Clearly, \( T(\mathbf{v}) \) is a linear combination of \( T(\mathbf{u}) \) and \( T^2(\mathbf{u}) \), so it remains within the span of \( \{ \mathbf{u}, T(\mathbf{u}), T^2(\mathbf{u}) \} \), i.e., within \( U \).

Show that U has dimension 3

To show that \( U \) has dimension 3, we need to demonstrate that \( \{ \mathbf{u}, T(\mathbf{u}), T^2(\mathbf{u}) \} \) are linearly independent.Suppose \( a\mathbf{u} + bT(\mathbf{u}) + cT^2(\mathbf{u}) = \mathbf{0} \). Applying \( T \) repeatedly, we get the system:1. \( a = 0 \) (since \( u eq \mathbf{0} \))2. \( b = 0 \) (since if \( T(\mathbf{u}) = \mathbf{0} \) and \( u eq \mathbf{0} \), then \( T^2(\mathbf{u}) = \mathbf{0} \))3. \( c = 0 \) (since if \( T^2(\mathbf{u}) = \mathbf{0} \))Since the only solution is \( a = b = c = 0 \), the vectors \( \mathbf{u}, T(\mathbf{u}), T^2(\mathbf{u}) \) are linearly independent. Hence, \( \dim(U) = 3 \).

Key concepts

Linear Transformations

Linear transformations are at the heart of linear algebra. They are functions that map elements from one vector space to another in a way that preserves the operations of vector addition and scalar multiplication.
Imagine a transformation like a machine where you put in a vector and get another one based on specific rules.
Some key aspects include:

• They are defined between vector spaces, such as from one space to itself, like in our problem with operator \( T\).
• They satisfy two properties: \( T(a \mathbf{v} + b \mathbf{w}) = aT(\mathbf{v}) + bT(\mathbf{w}) \) for all vectors \( \mathbf{v}, \mathbf{w} \) and scalars \( a, b \).
• Linear transformations can be represented by matrices, especially when finite-dimensional vector spaces are involved.

In the given exercise, the operator \( T \) repeats until it effectively nullifies the input, exemplified by \( T^3 = 0 \).
This indicates that the transformation imposes certain constraints on the vector space, which are foundational to showing properties like T-invariance.

Linear Independence

Understanding linear independence is crucial for determining the dimension of a subspace. Vectors are said to be linearly independent if no vector in the set can be written as a combination of others in that set.
Let's break it down more:

• If there exists a relation \( a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n = \mathbf{0} \) where all coefficients \( a_1, a_2, \ldots, a_n \) are zero, then the vectors are independent.
• Independence means that each vector adds a new dimension to the span, providing a unique direction.
• If vectors are dependent, you can remove one without changing the span.

In our context, checking linear independence of the set \( \{ \mathbf{u}, T(\mathbf{u}), T^2(\mathbf{u}) \} \) determined that no non-trivial combinations led to zero.
This confirms a full-dimensional subspace and ensures dimension 3, as shown by solving \( a = b = c = 0 \).

Vector Spaces

Vector spaces are collections of vectors that can be scaled and added together.
Think of them as foundational structures on which linear transformations operate.
Here are some notable points:

• They contain a zero vector (acting as the neutral element for addition).
• Any linear combination of vectors within a space remains in the space, displaying closure under addition and scalar multiplication.
• They have a structure defined by a set of axioms, ensuring predictability and consistency for operations.

In our exercise, the vector space \( V \) facilitates the application of the operator \( T \) and contains the subspace \( U \), which is span-based.
By exploring \( U \) and confirming it as T-invariant, we observe how transformations interact with subspaces, revealing deeper levels of structure and dimensionality beyond basic vector arrangements.