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

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

The matrix $$(A−λI)T(A - \lambda I)^{T}$$ is the same as the matrix $$(AT−λI)\left(A^{T} - \lambda I\right)$$, since the identity matrix is symmetric.
$$det(AT−λI)=det((A−λI)T)=det(A−λI)\det\left(A^{T} - \lambda I\right) = \det\left((A - \lambda I)^{T}\right) = \det (A - \lambda I)$$
From this it is obvious that the eigenvalues are the same for both $$AA$$ and $$ATA^{T}$$.