# Observations on the Computation of Eigenvalue and Eigenvector Jacobians +
## 6 Simplified Cases
### Real Symmetric Parent Matrix  
The first simplified case considered is when the parent matrix of the eigenvalues and eigenvectors is real and symmetric. Matrices of this structure frequently appear in practical problems from science and engineering (for instance, the Davenport solution to [PERSON]'s Problem [3]). In order to simplify and to parallel results from the existing literature, it is also necessary to make a choice for \(\mathbf{v}_{0}\); therefore, choose that \(\mathbf{v}_{0}=\mathbf{v}\) and \(\alpha=1\). Note that, as discussed previously, this is the choice usually made in practice, as it leads to the best condition for the calculation of the eigenvector derivative. To begin the simplifications for the symmetric case consider Equation (38), repeated here for convenience:  
\[\left[\mathbf{A}-\lambda\mathbf{I}-\mathbf{v}\mathbf{v}^{T}\mathbf{A}\right] \frac{\partial\mathbf{v}}{\partial\mathbf{A}_{\text{vec}}}=-\mathbf{v}^{T} \otimes\left(\mathbf{I}-\mathbf{v}\mathbf{v}^{T}\right). \tag{57}\]  
Note that the Hermitian transposes have been replaced by standard transposes, since the eigenvalues and eigenvectors are guaranteed to be real since \(\mathbf{A}\) is real and symmetric. In addition, note that \(\mathbf{v}_{0}\) has been replaced by \(\mathbf{v}\). Now, as discussed before, the coefficient matrix of the eigenvector Jacobian is full rank (and invertible) as long as the matrix \(\mathbf{A}\) is full rank and invertible. Making use of the fact that for an invertible matrix  
\[\mathbf{A}^{+}=\mathbf{A}^{-1}, \tag{58}\]  
where \(\mathbf{A}^{+}\) is the Moore-Penrose pseudoinverse of \(\mathbf{A}\)[43], it is possible to write  
\[\frac{\partial\mathbf{v}}{\partial\mathbf{A}_{\text{vec}}}=-\mathbf{v}^{T} \otimes\left[\left(\mathbf{A}-\lambda\mathbf{I}-\mathbf{v}\mathbf{v}^{T} \mathbf{A}\right)^{+}\left(\mathbf{I}-\mathbf{v}\mathbf{v}^{T}\right)\right]. \tag{59}\]From here, it is necessary to further consider the pseudoinverse term. Recognize that the pseudoinverse term in Equation (59) can be expressed as the addition of a matrix and an outer product  
\[\left(\mathbf{A}-\lambda\mathbf{I}-\mathbf{v}\mathbf{v}^{T}\mathbf{A}\right)^{+ }=\left(\mathbf{B}+\mathbf{c}\mathbf{d}^{T}\right)^{+}, \tag{60}\]  
where \(\mathbf{B}=\mathbf{A}-\lambda\mathbf{I}\), \(\mathbf{c}=-\mathbf{v}\), and \(\mathbf{d}=\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Using the case \(i\) identities presented in [44] when \(\mathbf{A}\) is real and symmetric, it can be shown that  
\[\left(\mathbf{A}-\lambda\mathbf{I}-\mathbf{v}\mathbf{v}^{T}\mathbf{A}\right)^ {+}=\left(\mathbf{A}-\lambda\mathbf{I}\right)^{+}-\lambda^{-1}\mathbf{v} \mathbf{v}^{T}. \tag{61}\]  
Substituting this result into Equation (59) yields  
\[\frac{\partial\mathbf{v}}{\partial\mathbf{A}_{\text{vec}}}=-\mathbf{v}^{T} \otimes\left(\left[\left(\mathbf{A}-\lambda\mathbf{I}\right)^{+}-\lambda^{-1} \mathbf{v}\mathbf{v}^{T}\right]\left[\mathbf{I}-\mathbf{v}\mathbf{v}^{T} \right]\right). \tag{62}\]  
Now, expanding the matrix multiplication in the Kronecker product gives  
\[\frac{\partial\mathbf{v}}{\partial\mathbf{A}_{\text{vec}}}=- \mathbf{v}^{T}\otimes\left[\left(\mathbf{A}-\lambda\mathbf{I}\right)^{+}- \lambda^{-1}\mathbf{v}\mathbf{v}^{T}\right.\] \[\left.\qquad\qquad\qquad\qquad-\left(\mathbf{A}-\lambda\mathbf{I }\right)^{+}\mathbf{v}\mathbf{v}^{T}+\lambda^{-1}\mathbf{v}\mathbf{v}^{T} \mathbf{v}\mathbf{v}^{T}\right], \tag{63}\]  
which, when taking into account that for symmetric matrices the pseudoinverse has the same null space as the matrix itself, simplifies to  
\[\frac{\partial\mathbf{v}}{\partial\mathbf{A}_{\text{vec}}}=-\mathbf{v}^{T} \otimes\left(\mathbf{A}-\lambda\mathbf{I}\right)^{+}, \tag{64}\]  
which is exactly the same result presented in [34]. With the simplified version of the eigenvector derivative in hand, the simplified eigenvalue Jacobian is trivial to find.