are both unitary and Hermitian (for $0 \le \theta \le 2\pi$). I call the latter type trivial, since its columns equal to plus/minus columns of the identity matrix. Do such matrices have any …

A stronger notion is unitary equivalence, i.e., similarity induced by a unitary transformation (since these are the isometric isomorphisms of Hilbert space), which again cannot happen between a …

Very good proof! However, an interesting thing is that you can perhaps stop at the third last step, because an equivalent condition of a unitary matrix is that its eigenvector lies on the unit circle, …

Definition (Unitary matrix). A unitary matrix is a square matrix $\mathbf {U} \in \mathbb {K}^ {n \times n}$ such that \begin {equation} \mathbf {U}^* \mathbf {U} = \mathbf {I} = \mathbf {U} …

In the case where H is acting on a finite dimensional vector space, you can essentially view it as a matrix, in which case (by for example the BCH formula) the relation you state in a) is valid. …

Even more, any unitary matrix is in fact a matrix exponential of some skew-Hermitian matrix. So it's no big surprise that their numbers of degrees of freedom are the same.

Unitary matrices are the complex versions, and they are the matrix representations of linear maps on complex vector spaces that preserve "complex distances". If you have a complex vector …

So I'm trying to understand why the columns of a unitary matrix form an orthonormal basis. I know it has something to do with the inner product, but I don't fully understand that either (we learned...

I am having trouble proving that Hermitian Matrices ($A = A^{*}$) are unitarily diagonalizable ($A = Q^{*}DQ$, where Q is a unitary matrix, $QQ^{*} = I$ and D is a ...

Of course, this argument proves directly that $U (n)$ is connected. Generalization: thanks to the $L^\infty$ functional calculus, we can prove the unitary group of a von Neumann algebra is …