Let $G=U(n)$ be the unitary group. We know that any unitary matrix is diagonalizable. Let $x$ be a unitary matrix. Then I've read the statement that any matrix $b$ commuting with $x$ is block-diagonal with blocks corresponding (in size) to the eigenspaces of $x$. I wonder how one should read this statement. I know that $b$ can be brought into this form by changing to an eigenbasis of $x$, but this does not imply that $b$ is block-diagonal in any basis, right? (I am afraid this is a stupid question)
Let $\lambda_1,...,\lambda_r$ be the distinct eigenvalues of $x$ with multiplicities $m_1,...,m_r$. Then I read that $dim(C_G(x)) = m_1^2 + \cdots + m_r^2$, where $C_G(x)$ is the centralizer of $x$ in $G$, i.e., the subgroup of matrices commuting with $x$. How to clarify this statement under consideration of my above worry? I understand that the formula comes from identifying $C_G(x)$ with unitary block matrices (which commute with $x$) where the blocks have the sizes $m_1,...,m_r$, but why can we do that?
I want to understand what it means for an element to be regular, which by definition means that the dimension of the centralizer is as small as possible. Here then $x$ regular means that the eigenspaces of $x$ are all one-dimensional. Why is this equivalent to: the centralizer $C(x)$ is conjugate to the group $T$ of all diagonal matrices in $G$? EDIT: I fail to see why the statement that there is a $g$ such that $gC(x)g^{-1} = T$ implies that $x$ has $n$ pairwise distinct eigenvalues.
A side question: is the centralizer always (for more general situations, e.g. in the algebraic group setting) a product of vector spaces?
That's a lot of questions in one, so let me begin with the first one: If $bx=xb$, then for any eigen-value $\lambda$ of $x$, consider the space $E = \ker(x-\lambda I)$. For any $v\in E$, $$ \lambda bv = b(xv) = x(bv) \Rightarrow bv \in E $$ Hence, $b(E) \subset E$. Also, since $x^{\ast} = x^{-1}$, it follows that $$ xb = bx \Rightarrow bx^{\ast} = x^{\ast}b \Rightarrow b^{\ast}x = xb^{\ast} $$ and hence, as above, $b^{\ast}(E) \subset E$, from which it follows that $$b(E^{\perp}) \subset E^{\perp}$$ Hence, we get the block-diagonal representation of $b$ as you have described.
I hope this helps to get you started.