Norm of an operator formed using a unitary operator

Question

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.

Answer 1

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 $1$.

Answer 2

No. Take the unitary $ A=\begin{pmatrix} 0&1\\ -1&0 \end{pmatrix} $ which satisfies your assumption. The matrix $B=\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$ has norm 1.

Answer 3

Here is the proof for $ ||B||\le 1$(This still doesn't answer the question):- Consider the vector space $ \mathbb{C}^n\otimes \mathbb{C}^n\otimes \mathbb{C}^n\otimes \mathbb{C}^n$. There consider the subspace spanned by $ e_i^{\otimes 4} $. Then $ e_i^{\otimes 4} $ forms a orthonormal basis for this subspace. And the matrix for $ A\otimes A\otimes A\otimes A$ restricted to this subspace wrt this basis is exactly $ B $. And we know that $ ||A^{\otimes 4}||=||A||^4=1 $. So $ ||B||\le 1$.