TITLE: Absolute value of coefficients of the characteristic polynomial of a unitary matrix
QUESTION [1 upvotes]: Let $A$ be a unitary $n\times n$ matrix. Show that $|c_k|\leq\binom{n}{k}$ for the coefficients of the characteristic polynomial of A, $\chi_A(X)=c_0+c_1X+\ldots +c_{n-1}X^{n-1}+X^n$.

I know that $\chi_A(X)$ factors completely into terms of degree 1 since $A$ has complex entries and that since $A$ is unitary, all eigenvalues have absolute value less that or equal to $1$. However, I don't know how I should proceed from there.

REPLY [0 votes]: So if tou know that $c_n=1$ then using Vieta's formulas you now that
$$\frac{c_{n-k}}{c_n}(-1)^k=\sum _{1\leq i_1<\cdots <i_k\leq n}\prod _{j=1}^k r_{i_j},$$
where $r_i$ are the roots of the polynomial. Take norm both sides, use that $c_n=1$ and use what you know, that $|r_i|\leq 1$ to conclude that $$|c_{n-k}|\leq |c_n|\sum _{1\leq i_1<\cdots <i_k\leq n}1=\binom{n}{k}.$$
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