Question

Exercise 9.3 .6 Show that the only subspaces of \(V\) that are \(T\) -invariant for every operator \(T: V \rightarrow V\) are 0 and \(V\). Assume that \(V\) is finite dimensional.

Short answer

The only \( T \)-invariant subspaces of \( V \) are \( \{ 0 \} \) and \( V \).

Step-by-step solution

Understanding Subspaces and T-invariance

A subspace of a vector space \( V \) is a subset that is closed under addition and scalar multiplication. A subspace \( W \subseteq V \) is said to be \( T \)-invariant if applying the linear operator \( T \) on any vector in \( W \) results in another vector that is still in \( W \). We need to show that only the trivial subspace \( \{ 0 \} \) and \( V \) itself are \( T \)-invariant for every operator \( T \).

Proving the Subspace \( \{ 0 \} \) is \( T \)-Invariant

Consider the subspace \( \{ 0 \} \). For any \( T: V \rightarrow V \), \( T(0) = 0 \). Hence, \( \{ 0 \} \) is \( T \)-invariant because applying \( T \) to any vector in \( \{ 0 \} \) results in the vector 0, which is still in \( \{ 0 \} \).

Proving the Subspace \( V \) is \( T \)-Invariant

Consider the subspace \( V \). For any linear operator \( T: V \rightarrow V \), every vector \( v \in V \) maps to another vector \( T(v) \in V \). This shows \( V \) is trivially \( T \)-invariant since nothing is mapped outside of \( V \).

Suppose there is a Non-Trivial \( T \)-Invariant Subspace

Assume there exists a non-trivial \( T \)-invariant subspace \( W \) such that \( \{ 0 \} \subset W \subset V \). By the definition of \( T \)-invariance, for any vector \( w \in W \) and any \( T \), \( T(w) \in W \).

Using the Arbitrary Nature of \( T \)

Since \( T \) is arbitrary and \( W \) is \( T \)-invariant for every \( T \), for any basis vector \( v_i \) of \( V \), we can construct a \( T \) such that \( T(v_i) ot\in W \) unless \( W = V \). Hence, a contradiction arises unless \( W = \{ 0 \} \) or \( W = V \).

Concluding the Proof

By utilizing the fact that for any assumed\( W \) being \( T \)-invariant makes \( W = \{ 0 \} \) or \( W = V \) due to our construction, we conclude that the only possible \( T \)-invariant subspaces are the trivial ones: \( \{ 0 \} \) and \( V \).

Key concepts

Subspaces

In the study of linear algebra, a subspace is essentially a space within a vector space itself. Let's break it down: A subspace of a vector space \( V \) is a smaller collection of vectors within \( V \) that maintains specific properties. These properties are:

• Closure under addition: If you take any two vectors in the subspace and add them together, the result is still in the subspace.
• Closure under scalar multiplication: If you multiply any vector in the subspace by a scalar (a constant), the resulting vector is still in the subspace.

For example, if \( V \) is a plane in three-dimensional space, then a line through the origin that lies in this plane is an example of a subspace. Subspaces are vital as they allow us to simplify complex problems by focusing on smaller, more manageable parts of the vector space.

Finite Dimensional Vector Spaces

A finite dimensional vector space is one that has a finite basis. To understand this, we need to know what a basis is. In simple terms, a basis of a vector space \( V \) is a set of vectors that are:

• Linearly independent: None of the vectors in the set can be formed by a combination of the others.
• Span \( V \): Every vector in \( V \) can be expressed as a linear combination of the basis vectors.

The dimension of \( V \) is the number of vectors in this basis. For instance, the standard three-dimensional space has a basis with three vectors, and so it is three-dimensional. This concept is crucial because it helps us understand the structure and properties of vector spaces. Finite dimensional vector spaces are particularly significant because they often pop up in practical applications, ranging from engineering to physics.

T-invariant Subspaces

A subspace \( W \) of a vector space \( V \) is \( T \)-invariant if it remains unchanged when a linear operator \( T \) is applied to any vector within this subspace. Essentially, applying \( T \) to a vector in \( W \) results in another vector that still belongs to \( W \).

Why is this concept important? \( T \)-invariant subspaces help us to simplify linear transformations by focusing on a smaller set of possibilities. In the context of finite dimensional vector spaces, it is fascinating to note that the only universally \( T \)-invariant subspaces for every operator \( T \) are the zero vector \( \{ 0 \} \) and \( V \) itself.

This means that no matter what transformation you apply, if a subspace remains invariant, it must essentially be either everything or nothing. This idea reduces complexity significantly in the realm of linear algebra and makes handling operators much more structured.