Chapter 7: Problem 3
A subspace \(Y\) of a normed space \(X\) is said to be invariant under a linear operator \(T: X \longrightarrow X\) if \(T(Y) \subset Y\). Show that an eigenspace of \(T\) is invariant under \(T\). Give examples.
Short Answer
Expert verified
An eigenspace of \( T \) is invariant under \( T \) since \( T(v) = \lambda v \) stays in the eigenspace.
Step by step solution
01
Understand the Problem
We need to show that the eigenspace corresponding to an eigenvalue \( \lambda \) of a linear operator \( T \) is a subspace of \( X \) that remains invariant under \( T \). To achieve this, we must show that if \( v \) is an eigenvector of \( T \) associated with \( \lambda \), then \( T(v) \) is also in the same eigenspace.
02
Define Eigenspace
Recall that for a linear operator \( T: X \to X \), the eigenspace associated with an eigenvalue \( \lambda \) is the set of all vectors \( v \) in \( X \) such that \( T(v) = \lambda v \). This set can be represented as \( E_{\lambda} = \{ v \in X : T(v) = \lambda v \} \).
03
Show Subspace is Invariant
Select any vector \( v \) in the eigenspace \( E_{\lambda} \). By definition of an eigenvector, \( T(v) = \lambda v \). The vector \( \lambda v \) is a scalar multiple of \( v \), which means \( \lambda v \) is still in \( E_{\lambda} \). Therefore, \( T(v) \in E_{\lambda} \), demonstrating that the eigenspace is invariant under \( T \).
04
Example of Invariance
Consider the operator \( T: \mathbb{R}^2 \to \mathbb{R}^2 \) defined by \( T(x, y) = (x, 0) \). The eigenspace for eigenvalue \( \lambda = 1 \) is all vectors of the form \( (x, 0) \), and \( T(x, 0) = (x, 0) \) shows it is invariant. This illustrates the concept of an eigenspace being invariant under \( T \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenspace
In functional analysis, an eigenspace is a fascinating concept that helps to understand how transformations affect vectors. When dealing with a linear operator, the eigenspace associated with a particular eigenvalue consists of all vectors that are transformed by this operator in a specific way. For a linear operator \( T: X \to X \), and a given eigenvalue \( \lambda \), the eigenspace \( E_{\lambda} \) is defined by
- \( E_{\lambda} = \{ v \in X : T(v) = \lambda v \} \)
Invariant Subspace
An invariant subspace in the context of a linear operator is extremely important to understand. If we have a linear operator \( T: X \to X \), a subspace \( Y \subset X \) is invariant under \( T \) if the image of \( Y \) under \( T \) is still entirely within \( Y \). In mathematical terms, this means
- \( T(Y) \subseteq Y \)
Linear Operator
A linear operator is essentially a function that maps vectors from one space to another while respecting the rules of vector addition and scalar multiplication. Formally, an operator \( T: X \to X \) is linear if, for any vectors \( u, v \in X \) and any scalars \( a, b \), it holds that
- \( T(au + bv) = aT(u) + bT(v) \)
Eigenvalue
The eigenvalue is a vital concept when exploring linear transformations. Given a linear operator \( T: X \to X \), an eigenvalue \( \lambda \) is a special type of scalar that allows the expression of the transformation of vectors in a neat form. Specifically, for a vector \( v \) that is not zero, if \( T(v) = \lambda v \), then \( \lambda \) is termed an eigenvalue of \( T \).
- The equation \( T(v) = \lambda v \) means \( v \) is simply a scaled version of itself under the action of \( T \).
Normed Space
A normed space is a type of vector space equipped with a norm, which allows the measurement of vector 'magnitude' or 'length.' For a vector space \( X \), the norm of a vector \( v \in X \) is typically represented by \( \|v\| \), and it must satisfy these properties:
- \( \|v\| \geq 0 \), and \( \|v\| = 0 \) only if \( v = 0 \)
- \( \|cv\| = |c| \cdot \|v\| \) for any scalar \( c \)
- \( \|u + v\| \leq \|u\| + \|v\| \)