Difference between revisions of "Inverse mapping"
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− | ''(inverse operator) of a single-valued onto mapping (operator) | + | ''(inverse operator) of a single-valued onto mapping (operator) $ f: M \to f[M] $'' |
− | + | A single-valued mapping $ g $ such that | |
+ | \begin{align} | ||
+ | g \circ f & = \mathsf{Id}_{X} \quad \text{on} ~ M, \qquad (1) \\ | ||
+ | f \circ g & = \mathsf{Id}_{Y} \quad \text{on} ~ f[M], \qquad (2) | ||
+ | \end{align} | ||
+ | where $ M \subseteq X $, $ f[M] \subseteq Y $, and $ X $ and $ Y $ are any sets. | ||
− | '' | + | If $ g $ satisfies only Condition (1), then it is called a '''left-inverse''' mapping of $ f $, and if it satisfies only Condition (2), then it is called a '''right-inverse''' mapping of $ f $. The inverse mapping $ f^{-1} $ exists if and only if for each $ y \in f[M] $, the inverse image $ {f^{\leftarrow}}[\{ y \}] $ consists of just a single element $ x \in M $. If $ f $ has an inverse mapping $ f^{-1} $, then the equation |
+ | $$ | ||
+ | f(x) = y \qquad (3) | ||
+ | $$ | ||
+ | has a unique solution for each $ y \in f[M] $. If only a right inverse $ f_{R}^{-1} $ exists, then a solution of (3) exists, but its uniqueness is an open question. If only a left inverse $ f_{L}^{-1} $ exists, then any solution is unique, assuming that it exists. | ||
− | A | + | If $ X $ and $ Y $ are vector spaces, and if $ A $ is a linear operator from $ X $ into $ Y $, then $ A^{-1} $ is also linear, if it exists. In general, if $ X $ and $ Y $ are endowed with some kind of structure, it may happen that certain properties of $ A $ are also inherited by $ A^{-1} $, assuming that it exists. For example: |
− | + | * If $ X $ and $ Y $ are Banach spaces, and if $ A: X \to Y $ is a closed operator, then $ A^{-1} $ is also closed. | |
+ | * If $ \mathcal{H} $ is a Hilbert space and $ A: \mathcal{H} \to \mathcal{H} $ is self-adjoint, then $ A^{-1} $ is also self-adjoint. | ||
+ | * If $ f: \mathbf{R} \to \mathbf{R} $ is an odd function, then $ f^{-1} $ is also odd. | ||
− | + | The continuity of $ A $ does not always imply the continuity of $ A^{-1} $ for many important classes of linear operators, e.g., completely-continuous operators. The following are important tests for the continuity of the inverse of a linear operator. | |
− | + | Let $ X $ be a finite-dimensional vector space, with a certain basis, and let $ A: X \to X $ be given by the matrix $ [a_{ij}] $ with respect to this basis. Then $ A^{-1} $ exists if and only if $ \det([a_{ij}]) \neq 0 $ (in this case, $ A $ and $ A^{-1} $ are automatically continuous). | |
− | + | Let $ X $ and $ Y $ be Banach spaces, and let $ A $ be a continuous linear operator from $ X $ into $ Y $. | |
− | + | # If $ \| A(x) \| \geq m \| x \| $, where $ m > 0 $, then $ A^{-1} $ exists and is continuous. | |
+ | # If $ X = Y $ and $ \| A \| < 1 $, then $ (\mathsf{Id} - A)^{-1} $ exists, is continuous, and $$ (\mathsf{Id} - A)^{-1} = \sum_{n = 0}^{\infty} A^{n}, $$ where the series on the right-hand side converges in the norm of the space $ \mathcal{L}(X) $. | ||
+ | # The operator $ A^{-1} $ exists and is continuous on all of $ Y $ if and only if the conjugate $ A^{*} $ has an inverse that is defined and continuous on $ X^{*} $. Here, $ (A^{-1})^{*} = (A^{*})^{-1} $. | ||
+ | # If $ A^{-1} $ exists and is continuous, and if $ \| A - B \| < \| A^{-1} \|^{-1} $, then $ B^{-1} $ also exists, is continuous, and $$ B^{-1} = A^{-1} \sum_{n = 0}^{\infty} [(A - B) A^{-1}]^{n}. $$ Therefore, the set of invertible operators is open in $ \mathcal{L}(X,Y) $ in the [[Uniform topology|uniform topology]] of this space. | ||
+ | # '''Banach’s Open-Mapping Theorem.''' If $ A $ is a one-to-one mapping of $ X $ onto $ Y $, then the inverse mapping, which exists, is continuous. This theorem has the following generalization: A one-to-one continuous linear mapping of a fully-complete space $ X $ onto a separated barrelled space $ Y $ is a topological [[Isomorphism|isomorphism]]. | ||
− | + | The spectral theory of linear operators on a Hilbert space contains a number of results on the existence and continuity of the inverse of a continuous linear operator. For example, if $ A $ is self-adjoint and $ \lambda $ is not real, then $ (A - \lambda \cdot \mathsf{Id})^{-1} $ exists and is continuous. | |
− | + | ====References==== | |
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− | |||
− | + | <table> | |
− | + | <TR><TD valign="top">[1]</TD><TD valign="top"> | |
− | + | N. Dunford, J.T. Schwartz, “Linear operators. General theory”, '''1''', Interscience (1958).</TD></TR> | |
− | + | <TR><TD valign="top">[2]</TD> <TD valign="top"> | |
− | + | L.V. Kantorovich, G.P. Akilov, “Functional analysis”, Pergamon (1982). (Translated from Russian)</TD></TR> | |
− | + | <TR><TD valign="top">[3]</TD><TD valign="top"> | |
− | + | W. Rudin, “Functional analysis”, McGraw-Hill (1979).</TD></TR> | |
− | + | <TR><TD valign="top">[4]</TD><TD valign="top"> | |
− | + | A.P. Robertson, W.S. Robertson, “Topological vector spaces”, Cambridge Univ. Press (1964).</TD></TR> | |
− | + | </table> | |
− | <table | ||
− | |||
− | |||
− | |||
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− |
Latest revision as of 03:29, 9 January 2017
(inverse operator) of a single-valued onto mapping (operator) $ f: M \to f[M] $
A single-valued mapping $ g $ such that \begin{align} g \circ f & = \mathsf{Id}_{X} \quad \text{on} ~ M, \qquad (1) \\ f \circ g & = \mathsf{Id}_{Y} \quad \text{on} ~ f[M], \qquad (2) \end{align} where $ M \subseteq X $, $ f[M] \subseteq Y $, and $ X $ and $ Y $ are any sets.
If $ g $ satisfies only Condition (1), then it is called a left-inverse mapping of $ f $, and if it satisfies only Condition (2), then it is called a right-inverse mapping of $ f $. The inverse mapping $ f^{-1} $ exists if and only if for each $ y \in f[M] $, the inverse image $ {f^{\leftarrow}}[\{ y \}] $ consists of just a single element $ x \in M $. If $ f $ has an inverse mapping $ f^{-1} $, then the equation $$ f(x) = y \qquad (3) $$ has a unique solution for each $ y \in f[M] $. If only a right inverse $ f_{R}^{-1} $ exists, then a solution of (3) exists, but its uniqueness is an open question. If only a left inverse $ f_{L}^{-1} $ exists, then any solution is unique, assuming that it exists.
If $ X $ and $ Y $ are vector spaces, and if $ A $ is a linear operator from $ X $ into $ Y $, then $ A^{-1} $ is also linear, if it exists. In general, if $ X $ and $ Y $ are endowed with some kind of structure, it may happen that certain properties of $ A $ are also inherited by $ A^{-1} $, assuming that it exists. For example:
- If $ X $ and $ Y $ are Banach spaces, and if $ A: X \to Y $ is a closed operator, then $ A^{-1} $ is also closed.
- If $ \mathcal{H} $ is a Hilbert space and $ A: \mathcal{H} \to \mathcal{H} $ is self-adjoint, then $ A^{-1} $ is also self-adjoint.
- If $ f: \mathbf{R} \to \mathbf{R} $ is an odd function, then $ f^{-1} $ is also odd.
The continuity of $ A $ does not always imply the continuity of $ A^{-1} $ for many important classes of linear operators, e.g., completely-continuous operators. The following are important tests for the continuity of the inverse of a linear operator.
Let $ X $ be a finite-dimensional vector space, with a certain basis, and let $ A: X \to X $ be given by the matrix $ [a_{ij}] $ with respect to this basis. Then $ A^{-1} $ exists if and only if $ \det([a_{ij}]) \neq 0 $ (in this case, $ A $ and $ A^{-1} $ are automatically continuous).
Let $ X $ and $ Y $ be Banach spaces, and let $ A $ be a continuous linear operator from $ X $ into $ Y $.
- If $ \| A(x) \| \geq m \| x \| $, where $ m > 0 $, then $ A^{-1} $ exists and is continuous.
- If $ X = Y $ and $ \| A \| < 1 $, then $ (\mathsf{Id} - A)^{-1} $ exists, is continuous, and $$ (\mathsf{Id} - A)^{-1} = \sum_{n = 0}^{\infty} A^{n}, $$ where the series on the right-hand side converges in the norm of the space $ \mathcal{L}(X) $.
- The operator $ A^{-1} $ exists and is continuous on all of $ Y $ if and only if the conjugate $ A^{*} $ has an inverse that is defined and continuous on $ X^{*} $. Here, $ (A^{-1})^{*} = (A^{*})^{-1} $.
- If $ A^{-1} $ exists and is continuous, and if $ \| A - B \| < \| A^{-1} \|^{-1} $, then $ B^{-1} $ also exists, is continuous, and $$ B^{-1} = A^{-1} \sum_{n = 0}^{\infty} [(A - B) A^{-1}]^{n}. $$ Therefore, the set of invertible operators is open in $ \mathcal{L}(X,Y) $ in the uniform topology of this space.
- Banach’s Open-Mapping Theorem. If $ A $ is a one-to-one mapping of $ X $ onto $ Y $, then the inverse mapping, which exists, is continuous. This theorem has the following generalization: A one-to-one continuous linear mapping of a fully-complete space $ X $ onto a separated barrelled space $ Y $ is a topological isomorphism.
The spectral theory of linear operators on a Hilbert space contains a number of results on the existence and continuity of the inverse of a continuous linear operator. For example, if $ A $ is self-adjoint and $ \lambda $ is not real, then $ (A - \lambda \cdot \mathsf{Id})^{-1} $ exists and is continuous.
References
[1] | N. Dunford, J.T. Schwartz, “Linear operators. General theory”, 1, Interscience (1958). |
[2] | L.V. Kantorovich, G.P. Akilov, “Functional analysis”, Pergamon (1982). (Translated from Russian) |
[3] | W. Rudin, “Functional analysis”, McGraw-Hill (1979). |
[4] | A.P. Robertson, W.S. Robertson, “Topological vector spaces”, Cambridge Univ. Press (1964). |
Inverse mapping. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Inverse_mapping&oldid=40159