Matrices and Unit Circle on an Orthonormal Basis

Question

This is a modification of my previous question from Treil's textbook:Unitary and Orthonormal.

Let $U: X\rightarrow X$ be an linear transformation on a inner product space. $U_n$ can be represented by a unit circle.

True or False: $U$ is unitary.

My guess is $U$ is not isometric then. Please correct me if I am wrong.

Answer 1

You already have a correct answer. But just in case you think that every example uses irrational numbers, here's one that doesn't:$$(x,y)\mapsto\left(x+\frac35y,\frac45y\right).$$A way of getting counter-examples as this one is this: take vectors $f_1,\ldots,f_n$ with norm $1$ such that at least two of them are not orthogonal. Then the linear map $U$ such$$(\forall k\in\{1,2,\ldots,n\}):U(e_k)=f_k,$$where $\{e_1,\ldots,e_n\}$ is the canonical basis, will be one such counter-example.

Answer 2

It is not true that $U$ is necessarily unitary. For instance, take $X = \Bbb C^2$ and $$ U = \pmatrix{1&1/\sqrt{2}\\0&1/\sqrt{2}} $$ Or, if you prefer the mapping notation, this is the transformation $f:(x,y) \mapsto (x + y/\sqrt{2},y/\sqrt{2})$.

This $U$ satisfies the condition with the basis $e_1 = (1,0), e_2 = (0,1)$. However, you may verify that $\|U(1,\sqrt{2} - 1)\| \neq \|(1,\sqrt{2} - 1)\|$.