Let $A$ be an $n \times n$ self adjoint matrix with eigen values $\lambda_1, \lambda_2, \cdots, \lambda_n$.
Let $\|X\|_2 = \sqrt {|x_1|^2 + |x_2|^2 + \cdots + |x_n|^2}$ for $X = (x_1,x_2, \cdots , x_n) \in \Bbb C^n$.
If $p(A) = a_0 I + a_1 A + a_2 A^2 + \cdots + a_n A^n$ then $\mathrm {sup}_{\|X\|_2 = 1}\ \|p(A)X\|_2$ is equal to
$1.\ \ \ \ \mathrm {max} \left \{a_0+ a_1 \lambda_j + a_2 {\lambda_j}^2 + \cdots + a_n {\lambda_j}^n\ :\ 1 \le j \le n \right \}$
$2.\ \ \ \ \mathrm {max} \left \{ \left |a_0+ a_1 \lambda_j + a_2 {\lambda_j}^2 + \cdots + a_n {\lambda_j}^n \right |\ :\ 1 \le j \le n \right \}$
$3.\ \ \ \ \mathrm {min} \left \{a_0+ a_1 \lambda_j + a_2 {\lambda_j}^2 + \cdots + a_n {\lambda_j}^n\ :\ 1 \le j \le n \right \}$
$4.\ \ \ \ \mathrm {min} \left \{ \left |a_0+ a_1 \lambda_j + a_2 {\lambda_j}^2 + \cdots + a_n {\lambda_j}^n \right |\ :\ 1 \le j \le n \right \}$
I find difficulty to solve it. Please help me.
Thank you in advance.
Number (2). The operator norm of a selfadjoint (or normal) matrix $A$ is $\|A\|=\max_j|\lambda_j|$. For any polynomial $p$, the eigenvalues of $p(A)$ are $\{ p(\lambda_j) \}_{j=1}^{n}$. So $\|p(A)\|=\max_{j}|p(\lambda_j)|$, if $A$ is a selfadjoint (or normal) matrix.
To prove this directly for this $A$, let $\{ e_n \}$ be an orthonormal basis of eigenvectors of $A$ with eigenvalues $\lambda_n$ (allow for repeated eigenvalues.) Then \begin{align} p(A)x & = p(A)\sum_{j=1}^{n}\langle x,e_j\rangle e_j\\ & = \sum_{j=1}^{n} p(\lambda_j)\langle x,e_j\rangle e_j. \\ \|p(A)x\|^2 & = \sum_{j=1}^{n}|p(\lambda_j)|^2 |\langle x,e_j\rangle|^2 \\ & \le \left(\max_{j}|p(\lambda_j)|\right)^2\sum_{j=1}^{n}|\langle x,e_j\rangle|^2 \\ & = \left(\max_{j}|p(\lambda_j)|\|x\|\right)^2 \end{align} From that $\|p(A)\| \le \max_j|p(\lambda_j)|$. The reverse inequality is obtained by choosing $x=e_j$ where $j$ is chosen to maximize $|\lambda_j|$. So $\|p(A)\|=\max_j|p(\lambda_j)|$.