Chapter 10: Problem 10
If a self-adjoint linear operator \(T: \mathscr{D}(T) \longrightarrow H\) is injective, show that (a) \(\overline{\mathscr{M}(T)}=H\) and \((b) T^{-1}\) is self-adjoint.
Short Answer
Expert verified
The closure of the range is \( H \) and the inverse is self-adjoint.
Step by step solution
01
Understand the Terms
Before solving the problem, it's crucial to understand the key terms in the question. A self-adjoint operator is one that satisfies \( T = T^* \), where \( T^* \) is the adjoint of \( T \). An operator is injective if it maps distinct elements to distinct elements. \( \overline{\mathscr{M}(T)} = H \) means the closure of the image of \( T \) is the whole space \( H \).
02
Show that \( \overline{\mathscr{M}(T)} = H \)
To demonstrate that \( \overline{\mathscr{M}(T)} = H \), we use the fact that \( T \) is injective and self-adjoint. An injective self-adjoint operator implies that the kernel is \( \{0\} \). The Hahn-Banach Theorem and the Riesz Representation Theorem ensure that the image of a dense subset under a self-adjoint operator is dense in \( H \). Therefore, the closure of the image \( \mathscr{M}(T) \) is \( H \).
03
Prove that \( T^{-1} \) is Self-adjoint
Since \( T \) is self-adjoint and \( \overline{\mathscr{M}(T)} = H \), \( T \) is also surjective. This allows us to construct the inverse \( T^{-1} \). To show that \( T^{-1} \) is self-adjoint, we note that self-adjoint operators that are bijective have self-adjoint inverses. Therefore, \( T^{-1} = (T^*)^{-1} = T^{-1}* \), which confirms that \( T^{-1} \) is self-adjoint.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Injective Operators
Injective operators are fundamental in understanding linear transformations. An operator, say \( T \), is injective if it has the property that it maps distinct elements from its domain \( \mathscr{D}(T) \) to distinct elements in the codomain \( H \). This implies that if \( T(x_1) = T(x_2) \), then necessarily \( x_1 = x_2 \). In other words, injectivity means there are no two different input values that result in the same output. This unique mapping feature makes injective operators one-to-one mappings.
- An important consequence of **injective** self-adjoint operators is having a trivial kernel, i.e., \( \text{kernel}(T) = \{ 0 \} \).
- Injective transformation ensures that no information is lost when an element of \( \mathscr{D}(T) \) is mapped to \( H \).
Closure of Image
The concept of *closure of the image* is pivotal in linear algebra, especially concerning operators. When we refer to the closure of the image \( \overline{\mathscr{M}(T)} \), we're looking at the smallest closed set containing the image of the operator \( T \). In the context of the problem, where \( T \) is self-adjoint and injective, the closure of its image spans the entire space \( H \) (i.e., \( \overline{\mathscr{M}(T)} = H \)).
- This indicates that the image is dense in \( H \), meaning for every point in \( H \), there is a sequence of image points that converge to it.
- Self-adjointness ensures that this occurs when the operator's kernel is 0, confirming functionality across the entire space of \( H \).
Inverse Operators
Inverse operators allow us to reverse the transformation applied by the original operator. For an injective and surjective operator \( T \) (which is bijective), an inverse operator \( T^{-1} \) exists. This means that for every element \( y \) in \( H \), there is a unique \( x \) in \( \mathscr{D}(T) \) such that \( T(x) = y \).
- The existence of \( T^{-1} \) implies \( T \) is both injective and surjective, effectively mapping \( H \) onto itself.
- If the operator \( T \) is self-adjoint, its inverse \( T^{-1} \) is also self-adjoint. This property arises from interacting with the commutativity of adjoint operations: \( T^{-1} = (T^*)^{-1} = T^{-1}* \).