/sci/ - Science & Math

>>14828948
>The spectral theorem feels like an overkill in this context. Surely there is some elementary proof of this fact?
Depends on how you define multiplicity of an eigenvalue for operators on a Hilbert space.
For finite dimensional spaces, we can take the determinant and look at the order of the zero to count algebraic multiplicities.
If you just mean the dimension of the kernel of [math](\lambda - U)[/math], then yeah, you can prove it by contradiction.
Decompose [math]\eta = u + v[/math], where [math]u \in Ker (\lambda - U)[/math] and [math]v \in Ker(\lambda - U)^{\perp}[/math]. [math]U[/math] fixes [math]Ker(\lambda - U)[/math] and [math]Ker ( \lambda - U)^{\perp}[/math], so you can construct an element from the kernel not in the span.
The argument is basically the spectral theorem argument but you have to fill in some gaps you prove while proving the spectral theorem.