Question is from a past QR Exam and is exactly as the title:
For which real, $n \times n$ symmetric matrices $A$ is $A-A^3$ invertible?
$A$ of course commutes with itself and the identity $I_n$, so we can write $A – A^3 = A(I_n + A)(I_n – A)$. So then for $A – A^3$ to be invertible, the matrix $A$ cannot have eigenvalues $\lambda = 0,\pm 1$. However, that seems to be independent of whether or not $A$ is symmetric. I must be missing something, because I don't see the connection with $A$ being symmetric.
Best Answer
Since $A$ is symmetric, it is diagonalizable and its eigenvalues are real. So, for some invertible matrix $P$, $PAP^{-1}=D$, for some diagonal matrix $D$, and $PA^3P^{-1}=D^3$. Therefore, $P(A-A^3)P^{-1}=D-D^3$ and, since $D$ is diagonal, $D-D^3$ is invertible if and only if none of the eigenvalues of $D$ is $0$, $1$, or $-1$.