A matrix and its transpose have the same set of eigenvalues/other version: AA and ATA^T have the same spectrum

Let $\sigma(A)$ be the set of all eigenvalues of $A$ . Show that $\sigma(A) = \sigma\left(A^T\right)$ where $A^T$ is the transpose matrix of $A$ .

Answer

The matrix $(A - \lambda I)^{T}$ is the same as the matrix $\left(A^{T} - \lambda I\right)$ , since the identity matrix is symmetric.

Thus:

$\det\left(A^{T} - \lambda I\right) = \det\left((A - \lambda I)^{T}\right) = \det (A - \lambda I)$

From this it is obvious that the eigenvalues are the same for both $A$ and $A^{T}$ .

Attribution
Source : Link , Question Author : Zizo , Answer Author : Invisible