91Ó°ÊÓ

In order to compare the eigenvalues of \(A\) and \(A^{T}\), we first need to find the eigenvalues of the matrices. Let \(\lambda\) be the eigenvalue of \(A\) and \(\tilde{\lambda}\) be the eigenvalue of \(A^{T}\). Then according to the definition of eigenvalues, we have: $$Ax = \lambda x$$ $$(A^{T})y = \tilde{\lambda} y$$ Where \(x\) and \(y\) are the eigenvectors associated with the eigenvalues.

In order to compare the eigenvalues, we need to look at the characteristic equation of the matrices. The characteristic equation of a matrix is given by: $$\text{det}(A - \lambda I)=0$$ and $$\text{det}(A^{T} - \tilde{\lambda} I)=0$$ Here, \(I\) is the identity matrix of size \(n \times n\).

Now, we need to derive the relationship between the determinant of a matrix and its transpose. The determinant of a transpose matrix is equal to the determinant of the matrix itself: $$\text{det}(A^{T}) = \text{det}(A)$$ Using this property, we can rewrite the characteristic equation of \(A^{T}\) as: $$\text{det}(A^{T} - \tilde{\lambda} I) = \text{det}((A - \tilde{\lambda} I)^{T}) = \text{det}(A - \tilde{\lambda} I)$$ From this, we can conclude that if \(\lambda\) is an eigenvalue of \(A\), then: $$\text{det}(A - \lambda I) = \text{det}(A^{T} - \lambda I) = 0$$ So, \(\lambda\) is also an eigenvalue of \(A^{T}\).

Since we have established that \(A\) and \(A^{T}\) have the same eigenvalues, we can now discuss the algebraic and geometric multiplicities. The algebraic multiplicity of an eigenvalue is the number of times it occurs as a root of the characteristic equation. Since the characteristic equations are the same, the algebraic multiplicities of the eigenvalues in \(A\) and \(A^{T}\) are also the same. The geometric multiplicity of an eigenvalue is the dimension of the eigenspace associated with it. For an eigenvalue \(\lambda\), the eigenspace associated with it is given by: $$\text{null}(A - \lambda I)$$ The geometric multiplicity of an eigenvalue is equal to the dimension of its eigenspace. Since we have established that \(A\) and \(A^{T}\) have the same eigenvalues and same characteristic equation, it follows that their eigenspaces are also identical, and hence the geometric multiplicities of their eigenvalues are also the same. In conclusion, the eigenvalues of a square matrix \(A\) and its transpose \(A^{T}\) are indeed the same, and they share the same algebraic and geometric multiplicities.