Mechanics of Continua and Structures
Definition A linear transformation from a Eucledian space $\mathbf{X}$ to a Eucledian space $\mathbf{Y}$ is a function
\(\begin{align} \boldsymbol{A} : \mathbf{X} \rightarrow \mathbf{Y} \\ \boldsymbol{x} \mapsto \boldsymbol{y} = \boldsymbol{A}\boldsymbol{x} \end{align}\) such that \(\begin{align} \boldsymbol{A}(\boldsymbol{x}_1 + \boldsymbol{x}_2) = \boldsymbol{A}\boldsymbol{x}_1 + \boldsymbol{A}\boldsymbol{x}_2 \end{align}\)
A matrix form for the linear transformation $\boldsymbol{A} : \mathbf{X} \rightarrow \mathbf{Y}$ is a matrix $A_{ij}$ that shows how basis elements $\boldsymbol{x}_j \in \mathbf{X}$ map to a linear combination of basis elements $\boldsymbol{y}_i \in \mathbf{Y}$ : \(\begin{align} \boldsymbol{x}_j \rightarrow \boldsymbol{A}\boldsymbol{x}_j = \sum_i A_{ij}\boldsymbol{y}_i \end{align}\)
Definition Given a linear transformation $\boldsymbol{A} : \mathbf{X} \rightarrow \mathbf{Y}$, then there exists a unique linear transformation (the adjoint) \(\begin{align} \boldsymbol{A}^T : \mathbf{X} \rightarrow \mathbf{Y} \end{align}\) that preserves the inner product: \(\begin{align} \langle \boldsymbol{y}, \boldsymbol{Ax} \rangle = \langle \boldsymbol{A}^T\boldsymbol{y}, \boldsymbol{x} \rangle \end{align}\) for all $\boldsymbol{x}$ and $\boldsymbol{y}$.
Note: Adjoint is independent of the choice of bases.
If the bases for $\mathbf{X}$ and $\mathbf{Y}$ are each orthonormal, then the matrix representation of the adjoint is the transpose of the matrix representation: \(\begin{align} \boldsymbol{A}^T\boldsymbol{y}_i = \sum_j A_{ij}\boldsymbol{x}_j \end{align}\)
To prove this relationship, we can verify the adjoint condition (6) for arbitary basis elements:
\[\begin{align} \langle \boldsymbol{A}^T\boldsymbol{y}_i,\boldsymbol{x}_j \rangle &= \langle \sum_k A_{ik}\boldsymbol{x}_k,\boldsymbol{x}_j \rangle \\ &= A_{ij} (\text{since the basis }\boldsymbol{x}_j \text{is orthonormal})\\ &= \langle \boldsymbol{y}_i,\sum_k A_{kj}\boldsymbol{y}_k \rangle (\text{since the basis }\boldsymbol{y}_i \text{is orthonormal}) \\ &= \langle \boldsymbol{y}_i,\boldsymbol{Ax}_j \rangle \end{align}\]Note: If the bases are not orthogonal,then the transpose of the matrix representation is not the matrix representation of the adjoint.
Any linear function $\boldsymbol{X} \rightarrow \mathbb{R}$ can be expressed as $\boldsymbol{x} \mapsto \langle \boldsymbol{y},\boldsymbol{x} \rangle $ for a unique $\boldsymbol{y}$.
For some $\boldsymbol{y} \in \mathbf{X}$, the adjoint of the linear function
\(\begin{align} \boldsymbol{y} : \mathbb{R} \mapsto \mathbf{X} \\ z \mapsto \boldsymbol{x} = z\boldsymbol{y} \end{align}\) is given by
\[\begin{align} \boldsymbol{y}^T : \mathbf{X} \rightarrow \mathbb{R}\\ \boldsymbol{x} \mapsto z = \langle \boldsymbol{y},\boldsymbol{x}\rangle \end{align}\]Verification that the adjoint condition given in (1) holds:
\[\begin{align} \langle \boldsymbol{x},\boldsymbol{y}z\rangle &= \langle \boldsymbol{x},\boldsymbol{y}\rangle z \\ &= \langle \boldsymbol{y},\boldsymbol{x}\rangle z \\ &= \langle \langle \boldsymbol{y},\boldsymbol{x} \rangle,z\rangle \\ &= \langle \boldsymbol{y}^T\boldsymbol{x},z \rangle \end{align}\]From above, we get
\[\begin{align} \boldsymbol{y}^T\boldsymbol{x} = \langle \boldsymbol{y},\boldsymbol{x} \rangle \end{align}\]Suppose \(\begin{align} \boldsymbol{y} = \sum_i y_i\boldsymbol{x}_i \end{align}\)
for a basis $\boldsymbol{x}_i$. Using the natural orthonormal basis 1 for $\mathbb{R}$, the matrix representation of the linear transformation $\boldsymbol{y}$ is : \(\begin{align} 1 \rightarrow \sum_i y_{i}\boldsymbol{x}_i, \end{align}\) such that the vector with components $y_i$ define the matrix representation.
Note: For the adjoint $\boldsymbol{y}^T$, the matrix representation is not the transpose of the vector with components $y_i$, unless the basis $\boldsymbol{x}_i$ is orthonormal.
The matrix representation of the adjoint is:
\[\begin{align} \boldsymbol{x}_j \mapsto \langle \boldsymbol{y}, \boldsymbol{x}_j \rangle \boldsymbol{1} = \langle \sum_i y_i \boldsymbol{x}_i,\boldsymbol{x}_j \rangle 1 = \sum_i \langle \boldsymbol{x}_i,\boldsymbol{x}_j \rangle y_i 1 \end{align}\]** For a non-orthonormal basis, the matrix representation of the adjoint is not $\boldsymbol{x}_j \mapsto y_j1$.