Let $N\in M(n, \mathbb{C})$ be an $n \times n$ normal matrix, meaning that $N^\ast N = NN^\ast$. The spectral theorem states that $N$ is unitarily diagonalizable
$$ N = U^{-1} \Lambda U,$$
where $U$ is a unitary matrix, and $\Lambda$ is a diagonal matrix containing complex numbers.
However, $M(n, \mathbb{C})$ is a $C^\ast$-algebra, where the involution is given by the conjugate transpose. This means that, for arbitrary $X\in M(n, \mathbb{C}),$
$$ ||X^\ast X|| = ||XX^\ast||.$$
In other words, within the norm function, arbitrary matrices behave like normal matrices. From this fact, is it possible to obtain a generalization of the spectral theorem or diagonalizability to arbitrary matrices $X \in M(n, \mathbb{C})$ within the norm function (something like: $||X|| = ||U^{-1} \Lambda U||$)? If so, does this also apply to arbitrary bounded linear operators on a Hilbert space?
Best Answer
Yes, even in infinite dimension, but in a completely useless way. Take any matrix $X$. Take $\Lambda$ be diagonal with $\|X\|$ in the 1,1 entry and zeroes elsewhere. Take $U=I$. Then $$\|X\|=\|U^*\Lambda U\|.$$ As opposed to the usual Spectral Theorem, this gives you exactly zero information about $X$.