Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 adj $\space A$ 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.


share|cite|improve this question
Most uninformative title. – Did Jan 18 '12 at 16:40
@Didier:I tried to make it more informative but $150$ is the limit :( – Quixotic Jan 18 '12 at 17:05
...The limit worries you because you put useless stuff in the title. Try something like Eigenvalues of adjoint of non-singular matrix. – Did Jan 18 '12 at 18:01
@Didier:That's cogent, i am putting it right now:) – Quixotic Jan 18 '12 at 18:28
up vote 6 down vote accepted

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 $A^{-1}$.

share|cite|improve this answer
This is more clear and neat, I got it just by reading once. Thanks :) – Quixotic Jan 18 '12 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.

share|cite|improve this answer
$^t\operatorname{adj} A$ 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 $ – Quixotic Jan 18 '12 at 15:52
What is your definition of $\operatorname{adj}(A)$? – Davide Giraudo Jan 18 '12 at 16:11
Same as here – Quixotic Jan 18 '12 at 16:22
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. – Davide Giraudo Jan 18 '12 at 21:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.