Characteristic polynomial for the composition of a linear transformation with itself

Question

Suppose I know the characteristic polynomial of the linear transformation $T$, can the characteristic polynomial of $L= T\circ T$ be obtained?

At the moment I don't know any theorem or result that can help me

Answer 1

In theory, yes. Let $n$ be the dimension of $T$. The characteristic polynomial of $T$ is $$\sum_{k=0}^n (-1)^k\sigma_k(\lambda_1,\ldots,\lambda_n)x^{n-k}$$ where $\lambda_1,\ldots,\lambda_n$ denote the eigenvalues of $T$ and $\sigma_k$ is the $k$-th elementary symmetric polynomial in the $\lambda_i$. The characteristic polynomial of $T^2$ is $$\sum_{k=0}^n (-1)^k \sigma_k(\lambda_1^2,\ldots,\lambda_n^2)x^{n-k}.$$ By the fundamental theorem of symmetric polynomials, each coefficient above is expressible as a polynomial in $\sigma_1(\lambda_1,\ldots,\lambda_n),\ldots,\sigma_n(\lambda_1,\ldots,\lambda_n)$. That is, the coefficients of the characteristic polynomial of $T^2$ can be obtained from the coefficients of the characteristic polynomial of $T$.

P.S. You might want to look into Newton's identities.