# Codigo Para Activar Maple 16 Please add link or file download options. You can also choose to Send by My Email instead. @Q: How is an orthogonal resolution number related to the rank of an operator? I’ve seen that if $A$ is a continuous linear operator on a finite dimensional inner product space then the number of nonzero eigenvalues of $A$ is bounded above by the rank of $A$. For example, if $U$ is an orthogonal resolution of the identity and if $\lambda$ is a nonzero eigenvalue of $A$ then you have $\lambda = \int_{\mathbb{R}} (A(\sqrt{U}) – \lambda \mathrm{id}) \, \mathrm{d}\mu$. But how can the rank of $A$ be smaller than the number of nonzero eigenvalues of $A$ when $U$ is not normal? A: To be sure, the rank of $A$ could be $0$. For example, if $A=\begin{pmatrix}0&0\\0&0\end{pmatrix}$, then $A$ has no eigenvalue. Also, the number of nonzero eigenvalues could be finite, not countable. On the other hand, by spectral theorem (see Schatten’s book, example 3.4.2), for any normal operator $A$ one has $R(\lambda,A)=\dim \ker(\lambda I-A)$ (these are the eigenvalues of $A$). Indeed, if $\dim \ker(\lambda I-A)=n$, and $v_1,\ldots,v_n$ are in $\ker(\lambda I-A)^\perp$, then $A v_i=\lambda v_i$ for all $i$. By normality, there are orthonormal vectors $w_1,\ldots,w_n$ such that $Av_i=\lambda v_i$ for all $i$. Therefore, $A^k v_i=(\lambda^k) v_i$ for all $i$ and all $k$ (note that $\lambda^k$ is an eigenvalue of $A^k$), i.e., $A^k v_i=(\lambda^k) v_i$ for all