Deciphering Notation for Linear Operators in Inner Product Spaces

AI Thread Summary
The discussion centers on understanding the notation in Theorem 6.11 regarding linear operators in inner product spaces. The left-hand side of the equation, (cT)*, represents the adjoint of the linear operator cT, which maps a vector v to c(T(v)). The right-hand side, denoted as \bar{c}T*, indicates the adjoint of the operator T multiplied by the complex conjugate of c. Participants clarify that the notation for linear transformations differs from that of matrices, emphasizing the importance of context in interpreting the adjoint operation. Ultimately, the confusion stemmed from a misunderstanding of the notation, which was resolved through discussion.
Defennder
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Homework Statement


This is from a linear algebra textbook I'm reading. I don't know whether there is universal agreement as to how notation should read for this, so this thread may well be meaningless. But I'll just post it to see if anyone can decipher what the question means. I'm asked to prove it, by the way, but I can't do so unless I understand the notation:

Theorem 6.11 Let V be an inner product space, and let T and U be linear operators on V. Then
(b)(cT)^* = \bar{c}T^* \ \mbox{for any c} \ \in F


Homework Equations


The Attempt at a Solution


I can understand the RHS of the question; that for a vector u in V, \bar{c}T^*(\vec{u}). The * here denotes adjoint of the linear operator T, but how do I interpret the LHS? It clearly cannot be cT^*(\vec{u}). Elsewhere, the book uses A* for a matrix A to denote the conjugate transpose of A, where every entry in A* is the complex transpose of the complex conjugate of the corresponding entry in A. But that can't be the interpretation either since if that is so, the LHS would be a row vector while the RHS a column vector.
 
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For the left hand side: define a new linear operator S := c T. Then the left hand side is S^*, so the adjoint of this new operator.

The proof is easy by the way, just writing out a string of identities which follow from known (defining) properties.
 
Defennder said:

Homework Statement


This is from a linear algebra textbook I'm reading. I don't know whether there is universal agreement as to how notation should read for this, so this thread may well be meaningless. But I'll just post it to see if anyone can decipher what the question means. I'm asked to prove it, by the way, but I can't do so unless I understand the notation:

Theorem 6.11 Let V be an inner product space, and let T and U be linear operators on V. Then
(b)(cT)^* = \bar{c}T^* \ \mbox{for any c} \ \in F


Homework Equations


The Attempt at a Solution


I can understand the RHS of the question; that for a vector u in V, \bar{c}T^*(\vec{u}). The * here denotes adjoint of the linear operator T, but how do I interpret the LHS? It clearly cannot be cT^*(\vec{u}). Elsewhere, the book uses A* for a matrix A to denote the conjugate transpose of A, where every entry in A* is the complex transpose of the complex conjugate of the corresponding entry in A. But that can't be the interpretation either since if that is so, the LHS would be a row vector while the RHS a column vector.
?T is a linear operator. cT is the linear operator that maps vector v into c(T(v)). (cT)* is the adjoint of that operator. You say yourself "The * here denotes adjoint of the linear operator T". The fact that "Elsewhere, the book uses A* for a matrix A to denote the conjugate transpose of A, where every entry in A* is the complex transpose of the complex conjugate of the corresponding entry in A. " is not relevant here because you are talking about linear transformations, not matrices. Use the interpretation that is relevant.

(Given a specific basis, a linear transformation can be written as a matrix. The two interpretations of "*" then coincide: the "adjoint of the linear transformation" has matrix (in that same basis) that is the "adjoint of the (original) matrix.)
 
I got it. The problem was that I didn't know how to interpret the notation. Thanks.
 
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