Let $P_A(x)$ denote the characteristic polynomial of a matrix $A_{n×n} , n \geq 2$. Let $P_A(x)-P_{A^{-1}}(x) = c$......(1), where $c$ is a constant. Then which type of matrix satisfy (1)
My Attempt:
I saw a matrix $$A = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} $$ which gives $P_A(x)-P_{A^{-1}}(x) = 0$.
I see here that $Det(A) = Det(A^{-1}) = 1$ and $Trace(A) = Trace (A^{-1})$ but I have no idea in general. Please help me. Thanks.
Since $P_M(x)=\det(xI-M)$ for any matrix $M$, we have the relation \begin{align*} P_A(x) &=\det(xI-A)\\ &=\det(A)\det(xA^{-1}-I)\\ &=\det(A)(-x)^n\det\left(\frac1xI-A^{-1}\right)\\ &=\det(A)(-x)^nP_{A^{-1}}(1/x). \end{align*} In other words, $$P_{A^{-1}}(x)=(-1)^n\frac{x^nP_A(1/x)}{\det(A)}.$$ The polynomial in the numerator represents "reversing" the polynomial $P_A$, by swapping its $x^0$ and $x^n$ coefficients, $x^1$ and $x^{n-1}$ coefficients, etc. Also, $(-1)^n\det A=P_A(0)$. Write $P(x)=P_A(x)$ for simplicity. We then suppose $P_A(x)-P_{A^{-1}}(x)=c$ for some constant $c$. This gives the identity $$\tag{1}P(x)-c=\frac{x^nP(1/x)}{P(0)}.$$ Since this also implies $$P(1/x)=c+\frac{x^{-n}P(x)}{P(0)},$$ we get $$P(x)=c+\frac{x^n}{P(0)}\left(\frac{x^{-n}P(x)}{P(0)}+c\right)=c\left(\frac{x^n}{P(0)}+1\right)+\frac{P(x)}{P(0)^2}.$$ If $P(0)=\pm 1$, then we also require $c=0$; otherwise, $$P(x)=\frac{c\left(\frac{x^n}{P(0)}+1\right)}{1-\frac1{P(0)^2}}.$$ We may substitute in $x=0$ to solve for $c$ and get $c=P(0)-1/P(0)$, and so $$P(x)=x^n+P(0),$$ which does satisfy $(1)$. In the case where $P(0)=\pm 1$ and $c=0$, any polynomial $$x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ with $a_0=\pm 1$ and $a_j=a_0a_{n-j}$ will satisfy $(1)$.
Characterizing the matrices with these characteristic polynomials will be a lot trickier, since a characteristic polynomial carries much less information than a matrix itself. You could write out the roots of such a polynomial and find all the possible Jordan normal forms, if you want, but there won't be a simple answer like "nilpotent" or "skew-symmetric."