Matrix norm of difference

Question

How would you go about calculating the norm \begin{align} \sup_\limits{v:\,||v||=1} ||(I_n-R_{\phi}\otimes I_{n-1})v|| \end{align} where $I_n$ is the $n$-dimensional identity and \begin{align} R_{\phi}=\begin{bmatrix} 1 & 0 \\ 0 & e^{i\phi} \end{bmatrix} \end{align} That difference should be a matrix with a whole lot of zeroes except at the indexes $n-1,n-1$ and $n,n$. $||\cdot||$ is the standard Euclidean norm $||a||=\sqrt{\sum_i a_i^2}$.

Answer 1

You can see $R_\phi\otimes I_{n-1}$ as the $(2n-2)\times(2n-2)$ diagonal matrix with diagonal $1,e^{i\phi}, 1,e^{i\phi},\cdots$. So $I_{2n-2}-R_\phi\otimes I_{n-1}$ is the diagonal matrix with diagonal entries $0,1-e^{i\phi}, 0,1-e^{i\phi},\cdots$.

The operator norm of a diagonal matrix is the maximum of the diagonal in absolute value. Thus $$ \|I_{2n-2}-R_\phi\otimes I_{n-1}\|=|1-e^{i\phi}|=2 (1-\cos \phi). $$