Eigenvalues of a Matrix and its Transpose are the Same

https://yutsumura.com/eigenvalues-of-a-matrix-and-its-transpose-are-the-same/

Recall that the eigenvalues of a matrix are roots of its characteristic polynomial.
Hence if the matrices A$A$ and AT$A^{\mathrm{T}}$ have the same characteristic polynomial, then they have the same eigenvalues.

So we show that the characteristic polynomial pA(t)=det(A−tI)$p_A(t)=det(A-tI)$ of A$A$ is the same as the characteristic polynomial pAT(t)=det(AT−tI)$p_{A^{\mathrm{T}}}(t)=det(A^{\mathrm{T}}-tI)$ of the transpose AT$A^{\mathrm{T}}$.

We have

pAT(t)=det(AT−tI)=det(AT−tIT)=det((A−tI)T)=det(A−tI)=pA(t).since IT=Isince det(BT)=det(B) for any square matrix B$$\begin{array}{rl}& p_{A^{\mathrm{T}}}(t)=det(A^{\mathrm{T}}-tI)\\ & =det(A^{\mathrm{T}}-t{I}^{\mathrm{T}})& & \text{since }{I}^{\mathrm{T}}=I\\ & =det\left((A-tI{)}^{\mathrm{T}}\right)\\ & =det(A-tI)& & \text{since }det(B^{\mathrm{T}})=det(B)\text{ for any square matrix }B\\ & =p_A(t).\end{array}$$

Therefore we obtain pAT(t)=pA(t)$p_{A^{\mathrm{T}}}(t)=p_A(t)$, and we conclude that the eigenvalues of A$A$ and AT$A^{\mathrm{T}}$ are the same.

Remark: Algebraic Multiplicities of Eigenvalues

Remark that since the characteristic polynomials of A$A$ and the transpose AT$A^{\mathrm{T}}$ are the same, it furthermore yields that the algebraic multiplicities of eigenvalues of A$A$ and AT$A^{\mathrm{T}}$ are the same.