# Characteristic polynomial of A, if $\det(\operatorname{adj}(\operatorname{adj}(A))) = 81$?

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

• Does $adj(A)$ mean the adjoint (so in the real case transpose)? – user190080 Jun 29 '15 at 18:48
• adjoint as in $Aadj(A)=det(A)I$ – Stabilo Jun 29 '15 at 18:50

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 $A$.
• so $n=3$ and $det(A)=3$... but how do I get the characteristic polynomial of A? – Stabilo Jun 29 '15 at 19:32
• $\det A = \lambda_1 \lambda_2 \lambda_3$ ($n=3$), and lambdas are integers; so, $\lambda_1 = 3$, $\lambda_2 \lambda_3 = 1$; $\lambda_2=\lambda_3=1$ or $\lambda_2=\lambda_3=-1$. – Michael Galuza Jun 29 '15 at 19:49
• @Anjan3 What if $n=2$? – Brian Fitzpatrick Jul 1 '15 at 19:31