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 !
$\newcommand{\norm}[1]{\left\|{#1}\right\|}$ 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.