TITLE: Eigenvalues of adjoint of non-singular matrix
QUESTION [5 upvotes]: How could we prove that

If $\lambda_1,\lambda_2,\lambda_3,\cdots \lambda_n $ are the
eigenvalues of a non-singular square matrix $A$ then eigenvalues of $~\text{adj} 
 A~$ are $~\frac {\det A}{\lambda_1},\frac {\det A}{\lambda_2},\frac
 {\det A}{\lambda_3},\cdots \frac {\det A}{\lambda_n}$.

I stumbled upon this property while solving a MCQ type question, in the solution there is no proof, I was just wondering if anybody could show me how to prove this one.
Thanks,

REPLY [10 votes]: As Davide answer shows, using the identity $adj(A)=\det(A)A^{-1}$ this problem can be reduced to showing that the eigenvalues of $A^{-1}$ are exactly the inverses of the ones of $A$.
This is intuitively obvious, since $Ax=\lambda x \Rightarrow \frac{1}{\lambda}x = A^{-1}x$, but there could be issues with the multiplicities.
To formally prove it, note that
$$\det(\lambda I -A^{-1}) = \frac{\det(A) \det(\lambda I -A^{-1})}{\det(A)}= \frac{\lambda^n \det(A- \frac{1}{\lambda}I)}{\det(A)} \,.$$
This way you can relate the characteristic polynomials of $A$ and $A^{-1}$.