Find $\|\cdot\|_2$ norm of block matrix 
Compute the square norm $\|\cdot\|_2$ of the following block matrix
  \begin{bmatrix}
     O & I_n & \dots  & O \\
     I_n & O & \ddots  & O \\
     \vdots & \ddots & \ddots & I_n \\
     O & O & I_n  & O
\end{bmatrix} 
  where $I_n$ is the identity matrix of order $n$.

I have no idea to find norm of the above block matrix. Can anyone give me a hint? Thank you in advance !
 A: $\newcommand{\norm}[1]{\left\|{#1}\right\|}$
First of all you have to know or demonstrate that
$$\norm{S \otimes T} \leq \norm{S}\norm{T}$$
Then notice that the matrix $M$ you address is
$$M=J \otimes I$$
where $I$ is the identity matrix and $J$ the generalization of this matrix
$$J = \left[\begin{array}{cccc}
0 & 1 & 0 & 0\\
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 0\\ 
\end{array}\right ]$$
then you have
$$\norm{J \otimes I} \leq \norm{J}\norm{I}$$
i.e.
$$\norm{J \otimes I} \leq \norm{J}$$
Now you have to find the norm of J. There could be an easier way, but I'm not seeing other thing that diagonalize the matrix which is symmetric.
I would consider the characteristic polynomial. Considering the resolvant matrix:
$$J- \lambda I = \left[\begin{array}{cccc}
\lambda & 1 & 0 & 0\\
1 & \lambda & 1 & 0\\
0 & 1 & \lambda & 1\\
0 & 0 & 1 & \lambda\\ 
\end{array}\right ]$$
which is ruled by the following succession (cfr. https://en.wikipedia.org/wiki/Tridiagonal_matrix):
$$ p_0(\lambda)=1 $$
$$ p_1(\lambda)=\lambda $$
$$ p_n(\lambda)=\lambda p_{n-1}-p_{n-2} $$
So for every $n$ you can find the eigenvalues you need to diagonalize and then get the max which is your norm.
Being more specific as user1551 pointed out this is a special kind of tridiagonal Toeplitz Matrix and the eigen values are given by 
$$ \lambda_k=2 \cos(k \pi)/ (n+1) $$
with $k=1..n$ so I guess you also have an explicit formula for your norm.
