How could we prove that

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

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.

  • 3
    $\begingroup$ Most uninformative title. $\endgroup$
    – Did
    Jan 18, 2012 at 16:40
  • $\begingroup$ @Didier:I tried to make it more informative but $150$ is the limit :( $\endgroup$
    – Quixotic
    Jan 18, 2012 at 17:05
  • $\begingroup$ @Didier:That's cogent, i am putting it right now:) $\endgroup$
    – Quixotic
    Jan 18, 2012 at 18:28
  • 2
    $\begingroup$ This is not the adjoint, it is the adjugate. These are different $\endgroup$
    – Nick Alger
    May 8, 2021 at 3:49

2 Answers 2


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 $A^{-1}$.

  • $\begingroup$ This is more clear and neat, I got it just by reading once. Thanks :) $\endgroup$
    – Quixotic
    Jan 18, 2012 at 17:56

The key is the identity $\operatorname{adj} A\cdot A=\det A \cdot I_n$, and since $A$ is not singular we have $\operatorname{adj} A=\det(A)\cdot A^{-1}$. The eigenvalues of $A^{-1}$ are the respective inverses of the eigenvalues of $A$ with the same algebraic multiplicity as @N. S. showed.

  • $\begingroup$ $^t\operatorname{adj} A$ does this means transpose of $\operatorname{adj} A$? But then I am only aware of the identity $\operatorname{adj} A\cdot A= A \cdot \operatorname{adj} A=\det A \cdot I_n $ $\endgroup$
    – Quixotic
    Jan 18, 2012 at 15:52
  • $\begingroup$ What is your definition of $\operatorname{adj}(A)$? $\endgroup$ Jan 18, 2012 at 16:11
  • $\begingroup$ Same as here $\endgroup$
    – Quixotic
    Jan 18, 2012 at 16:22
  • $\begingroup$ Ok, what I took for adjoint is in fact the matrix $C$. So I will edit my answer in order to be conform with your link. $\endgroup$ Jan 18, 2012 at 21:36

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