Bounding a Symmetric Matrix Consider the following $n \times n$ matrix $A$, which has 1's on the superdiagonal and subdiagonal and 0's elsewhere, i.e.
$$\begin{pmatrix}
0 & 1 & 0 & \cdots & \cdots & \cdots & \cdots & 0\\
1 & 0 & 1 & 0 & & & & \vdots\\
0 & 1 & 0 & 1 & \ddots & & & \vdots\\
\vdots & 0 & \ddots & \ddots & \ddots & \ddots & & \vdots\\
\vdots & & \ddots & \ddots & \ddots & \ddots & 0 & \vdots\\
\vdots & & & \ddots & 1 & 0 & 1 & 0\\
\vdots & & & & 0 & 1 & 0 & 1\\
0 & \cdots & \cdots  & \cdots & \cdots & 0 & 1 & 0\\
\end{pmatrix}$$
I want to show that the operator (spectral) norm of $A$ is bounded from above by $2$, namely $||A|| \leq 2$. 
In trying to prove this statement I started by decomposing $A$ into a sum of two matrices $A_{\mbox{super}}$ and $A_{\mbox{sub}}$, where $A_{\mbox{super}}$ has 1's on the superdiagonal and 0's elsewhere. Similarly, I define $A_{\mbox{sub}}$to have 1's on the subdiagonal and 0's elsewhere. Thus, I write the norm of $A$ as
$$||A||= ||A_{\mbox{super}} + A_{\mbox{sub}}||$$
and then use the triangle equality to bound the norm from above, namely 
$$||A_{\mbox{super}} + A_{\mbox{sub}}|| \leq ||A_{\mbox{super}}|| + ||A_{\mbox{sub}}||$$
It is at this point that I am unsure on how to proceed. My surmise would be to bound $A_{\mbox{super}}$ by 1 and similarly bound $A_{\mbox{sub}}$ by 1. But am not sure how this would work out. 

If you are talking about basic standard matrix norm which is also equivalent to spectral norm, then just compute $A_{sub}^TA_{sub}$ and find the maximum absolute value eigenvalue of that, and take square root, which is equivalent to the basic standard matrix norm. You should find that the product matrix is all $1$ on the diagonal except for one zero, and everywhere else zero, so the maximum absolute value of eigenvalue is $1$ and you are done.
