$\lambda^3 A x + \lambda B x = C x$ Suppose $A,B,C$ are real $n\times n$ matrices such that $\det(A) \neq 0$. How can I prove that whenever $n$ is odd, there exist $\lambda$ and $x\neq 0$ which satisfy $\lambda^3 A x + \lambda B x = C x$? Is there some intuition which would help to find an example in 2 dimensions for which no $\lambda$ and $x\neq 0$ exist?
 A: Suppose first that $A=\operatorname{Id}$. Consider the matrix $\lambda^3\operatorname{Id}-\lambda B-C$. Its determinant is a polynomial whose degree is $3n$; not just $\leqslant3n$, but $3n$. That's because each entry of the main diagonal of $\lambda^3\operatorname{Id}-\lambda B-C$ is a polynomial of degree $3$ and all other entries of the matrix are polynomials whose degree is $\leqslant1$. Since the degree is odd, there is a $\lambda\in\mathbb R$ for which the determinant is $0$. Therefore, there is a non-zero vector $x$ such that $\lambda^3x+\lambda Bx-Cx=0$.
In the general case, just multiply everything by $A^{-1}$ and you'll get this case.
A: The determinant polynomial equation $\det(\lambda^3 A +\lambda B - C)=0$ also gives a counterexample for $n=2$. Take $A=I_2$, $B=0$ and
$$
C=\begin{pmatrix} 1 & -2 \cr 1 & -1\end{pmatrix}.
$$
Then $\det(\lambda^3 A +\lambda B - C)=0$ says that $\lambda^6+1=0$, which has no real solution. Indeed, there is no $x\neq 0$ and no real $\lambda$ such that $\lambda^3 A x + \lambda B x = C x$.
A: For a fixed $\lambda$, there is $x\ne 0$ satisfying $\lambda^3Ax + \lambda Bx = Cx$ if and only if $\det(\lambda^3 A +\lambda B - C)=0$. Now the determinant of $\lambda^3 A + \lambda B - C$ will be a polynomial in $\lambda$ with the coefficient of $\lambda^{3n}$ being $\det(A)$ (one can check this from the permutation definition of the determinant). Therefore since $\det(A)\ne 0$, the determinant of $\lambda^3 A + \lambda B -C$ will be a polynomial of odd degree in $\lambda$, and hence there will be some real value of $\lambda$ making the determinant 0 (and thus some $x\ne 0$ such that $\lambda^3Ax + \lambda Bx = Cx$).
