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Chapter 7: Problem 54

Prove that \(A\) and \(A^{T}\) have the same eigenvalues. Are the eigenspaces the same?

Short Answer

Expert verified
Yes, matrix \(A\) and its transpose \(A^{T}\) have the same eigenvalues. However, the eigenspaces may not be the same.

Step by step solution

01

Understanding Eigenvalues

Eigenvalues are defined as the scalars \(\lambda\) which satisfy the characteristic equation \(\text{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix. For a matrix and its transpose, the determinant would be the same, hence the eigenvalues will be equal.
02

Prove Eigenvalues are Same

The eigenvalues of a matrix \(A\) are found by solving the characteristic equation \(\text{det}(A - \lambda I) = 0\). But the determinant of a matrix doesn't change upon transposition, therefore we can conclude that \(\text{det}(A - \lambda I) = \text{det}(A^{T} - \lambda I)\). This implies that \(A\) and \(A^{T}\) have the same eigenvalues.
03

Understanding Eigenspaces

Eigenspaces correspond to each distinct eigenvalue and consist of the eigenvectors associated with that eigenvalue and the zero vector. That means eigenvectors for a matrix and its transpose might not be the same, even when the eigenvalues are equal.
04

Discussing if Eigenspaces are Same

While \(A\) and \(A^{T}\) have the same eigenvalues, they do not necessarily have the same eigenvectors, implying the eigenspaces may indeed be different. This is because the eigenvectors are found by plugging in the eigenvalues into \((A - \lambda I)X = 0\) or \((A^{T} - \lambda I)X = 0\) where \(X\) is an eigenvector, and since \(A\) and \(A^{T}\) might not be the same, the resulting \(X\) (eigenvector) might not be the same.

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