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Let $A$ be a square real matrix whose eigenvalues are positive integers, with

$$\det(\operatorname{adj}(\operatorname{adj}(A))) = 81 \, .$$

What is the characteristic polynomial of A?

Any hints?

Thanks

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  • $\begingroup$ Does $adj(A)$ mean the adjoint (so in the real case transpose)? $\endgroup$ – user190080 Jun 29 '15 at 18:48
  • $\begingroup$ adjoint as in $Aadj(A)=det(A)I$ $\endgroup$ – Stabilo Jun 29 '15 at 18:50
  • $\begingroup$ adj = adjugate (!= adjoint) $\endgroup$ – Andreas H. Jun 29 '15 at 22:01
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Hint. For a $j\times j$ matrix $M$ we have $$ \det\big(\DeclareMathOperator{adj}{adj}\adj(M)\big)=\det(M)^{j-1} $$

Suppose $A$ is $n\times n$. Then $$ \det\big(\adj\big(\adj(A)\big)\big) = \det\big(\adj(A)\big)^{n-1} = \det(A)^{(n-1)^2} $$

Now, since the determinant of $A$ is the product of the eigenvalues of $A$ we know that $\det(A)$ is a positive integer. Thus $n$ satisfies $$ \det(A)=81^{1/(n-1)^2}\in\Bbb N $$ Finally, note that the characteristic polynomial is $$ \chi_A(t)=(t-\lambda_1)\dotsb(t-\lambda_n) $$ where $\lambda_1,\dotsc,\lambda_n$ are the eigenvalues of $A$.

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  • $\begingroup$ so $n=3$ and $det(A)=3$... but how do I get the characteristic polynomial of A? $\endgroup$ – Stabilo Jun 29 '15 at 19:32
  • $\begingroup$ $\det A = \lambda_1 \lambda_2 \lambda_3$ ($n=3$), and lambdas are integers; so, $\lambda_1 = 3$, $\lambda_2 \lambda_3 = 1$; $\lambda_2=\lambda_3=1$ or $\lambda_2=\lambda_3=-1$. $\endgroup$ – Michael Galuza Jun 29 '15 at 19:49
  • $\begingroup$ Yes, I saw that, thanks. They are positive too, so.. 3,1,1 $\endgroup$ – Stabilo Jun 29 '15 at 19:53
  • $\begingroup$ So the discussion says that such characteristic polynomial may be not unique $\endgroup$ – Anjan3 Jul 1 '15 at 9:26
  • $\begingroup$ @Anjan3 What if $n=2$? $\endgroup$ – Brian Fitzpatrick Jul 1 '15 at 19:31

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