A square matrix $A$ of order $3$ with complex entries is such that $Tr(A)=Tr(A^2)=0$ then find the value of $|A^2+I_3|-|A^2|$.
My attempt:
Let $A$ have eigen values $a,b,c$, so $A^2$ has eigen values $a^2,b^2,c^2$, so we have $a+b+c=0,a^2+b^2+c^2=0$ so we get $ab+bc+ca=0$ so the characteristic equation of $A$ is $A^3-abcI=O$. Let $A^2+I$ have eigen values $a’,b’,c’$ so $a’+b’+c’=3$. Now I’m stuck. I am not sure whether I’m going in the right direction or not but I’ve written whatever I thought that may be useful. Please help to approach further.
Hello :) your attempt is correct.
From $a=-(b+c) $ follows $a^2=b^2+2bc+c^2$. On the other hand $a^2=-(b^2+c^2) $. This leads to $b^2+bc+c^2=0$. Hence, $b=c\cdot e^{\frac{2\pi}{3}i}$ or $b=c\cdot e^{-\frac{2\pi}{3}i}$. It means there is a $z\in\mathbb C$ with $(a, b, c) =(e^{\frac{2\pi}{3}i} z, e^{-\frac{2\pi}{3}i}z, z) $ or $(a, b, c) =(e^{-\frac{2\pi}{3}i} z, e^{\frac{2\pi}{3}i}z, z) $.
The eigenvalues of $A^2+I$ are $a^2+1, b^2+1$ and $c^2+1$. Hence, we get $$\det(A^2+1)-\det(A^2)=(a^2+1) (b^2+1) (c^2+1)-a^2b^2c^2.$$ Can you finish from here?