Let $x = \begin{bmatrix} x_1 & x_2 \end{bmatrix}^T$ where $x_1,x_2 \in \mathbb{R}^n$, $Ax=A_1x_1+A_2x_2$. Based on the largest eigenvectors (and their corresponding eigenvalues) $v_i : \frac{\|A_iv_i\|_2}{\|v_i\|_2}=\|A_i\|_2=\lambda_i$ is it possible to calculate $\|A\|_2$?
At first I thought $\|A\|_2=\max(\lambda_1, \lambda_2)$, but this is not the case:
$\|A\|_2=\max_w\frac{\|\sum A_iw_i\|_2}{\|w\|_2}=\left(\max_w\frac{\|\sum A_iw_i\|_2^2}{\|w\|_2^2}\right)^{\frac{1}{2}}=\left(\max_w\frac{\|\sum A_iw_i\|_2^2}{\sum\|w_i\|_2^2}\right)^{\frac{1}{2}}\neq \left(\max_w\frac{\sum\| A_iw_i\|_2^2}{\sum\|w_i\|_2^2}\right)^{\frac{1}{2}}$
Can $\|A\|_2$ be expressed using only $v_1$, $v_2$, $\lambda_1$ and $\lambda_2$? If not, I am certain at least an upper bound can be calculated, that is $\|A\|_2 \leq (\lambda_1^2+\lambda_2^2)^{\frac{1}{2}}$; can a better upper bound be calculated as a function of $v_1$, $v_2$?
The point is to find a formula that does not contain any matrix multiplication.
Since $A = \begin{bmatrix} A_1 & A_2 \end{bmatrix}$, then
$$\left\|A\right\|_2 = \sqrt{\lambda_{\max}(AA^T)} = \sqrt{\lambda_{\max}\left(A_1A_1^T + A_2A_2^T\right)}$$