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Every linear map between finite-dimensional inner product spaces has a unique adjoint
Statement
Every linear map between finite-dimensional real or complex inner product spaces has a unique adjoint .
Facts & Assumptions
Given: A linear map between finite-dimensional inner product spaces.
Every linear functional on a finite-dimensional inner product space has a unique representing vector (Finite-dimensional Riesz representation: every functional is uniquely ).
An adjoint must satisfy for all (The adjoint is characterised by ).
Inner products separate vectors: equality of all pairings forces equality of the paired vectors (Inner products separate vectors, and the induced norm is homogeneous: ).
Proof
Fix . The function is a linear functional on , so [L1] supplies a unique vector, call it , satisfying [L2].
For scalars and , pairing the representatives from step 1.1 shows and have the same pairing with every . By [L3] they are equal. Thus is linear.
Any adjoint must assign to each the unique representative from step 1.1, so it equals . This proves existence and uniqueness, including when either space is zero.
Depends on
- The adjoint $T^*:W\to V$ is characterised by $\langle Tv,w\rangle_W=\langle v,T^*w\rangle_V$
- Finite-dimensional Riesz representation: every functional is uniquely $v\mapsto\langle v,w\rangle$
- Inner products separate vectors, and the induced norm is homogeneous: $\lVert\lambda v\rVert=|\lambda|\lVert v\rVert$
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 42 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- Sheldon Axler, Linear Algebra Done Right, 4th ed., discussion following definition 7.1 (standard reference, not scraped)
- Sergei Treil, Linear Algebra Done Wrong, §5.5.1 (standard reference, not scraped)