Suppose, $ A $ is a unitary matrix in $ M\_n(\mathbb{C}) $ given by $ (a\_{i,j})\_{1\le i,j\le n} $ which has the property that, for all the basis elements $ e\_i $, $ Ae\_i\ne |\lambda| e\_j $ for all $i,j $ and $ |\lambda|=1 $. Then consider the matrix $ B=(a\_{i,j}^4)\_{1\le i,j\le n} $. Then is it true that $ ||B||< 1 $?


I could prove that, $ ||B||\le 1 $, but I want to know if its strictly less than 1 in this case.


Also is it possible to get such a matrix $ A $ for which $ ||B||=1 $ and $ B $ actually attends its norm when $ A $ is a operator on a infinite dimensional Hilbert space with basis $ \{e\_i\}\_i $?


Edit: Will Sawin already posted a solution for the finite dimensional case. So I just edited the infinite dimensional case and added a condition I forgot to mention earlier.



Norm of an operator formed using a unitary operator
I will show $||B||<1$ (in the finite-dimensional case). Suppose $|Bv|=|v|$ for some nonzero $v$. Then as $|Bv|$ is the projection of $A^{\otimes 4} ( \sum\_i v\_i e\_i^{\otimes 4})$ onto the subspace generated by $e\_i^{\otimes 4}$, and $|A^{\otimes 4} ( \sum\_i v\_i e\_i^{\otimes 4})|=| \sum\_i v\_i e\_i^{\otimes 4}|=|v|$, it follows that $A^{\otimes 4} ( \sum\_i v\_i e\_i^{\otimes 4})$ lies in the space generated by $e\_i^{\otimes 4}$. 


In other words, for each $j\_1,j\_2,j\_3,j\_4$ not all equal, we have $\sum\_i A\_{ij\_1} A\_{ij\_2} A\_{ij\_3} A\_{ij\_4} v\_i =0$. Now suppose $j\_1,j\_2,j\_3$ are not all equal, then $j\_4$ can be anything and this equation will still apply. Because $A\_{ij\_4}$ for different values of $j\_4$ form a basis, it follows that for all $i$, for all $j\_1,j\_2,j\_3$ not all equal. $A\_{ij\_1} A\_{ij\_2} A\_{ij\_3} v\_i=0$. Take some $i$ such that $v\_i$ is not zero. By assumption on $A$ there are $j$ and $k$ such that neither $A\_{ij}$ nor $A\_{ik}$ vanishes. Taking $j\_1=j,j\_2=k,j\_3=k$ we get a contradiction. So $|Bv|<|v|$ for all nonzero $v$ and the norm is less than $1$.


In the infinite-dimensional case, it can equal exactly one. Consider an operator on $l^2$ given by an infinite block-diagonal matrix with two-by-two blocks, each a two-by-two unitary matrix, and converging to a diagonal matrix but never actually reaching it. Then the fourth power will have blocks converging to a two-by-two unitary diagonal matrix, hence with norm converging to $1$, so the fourth power has norm $1$.


