bounded spectral norm of $\left(\frac{1}{n} A A^{H}-z I_{m \times m}\right)^{-1}$ for any complex-valued with a nonzero imaginary part.

Question

Let $\boldsymbol{A} \in \mathbb{C}^{m \times n}, m \geq n, \operatorname{rank}\{\boldsymbol{A}\}=n$

I want to Show that for all sizes the matrix $\left(\frac{1}{n} A A^{H}-z I_{m \times m}\right)^{-1}$ has a bounded spectral norm for any complex-valued $z$ with a nonzero imaginary part.

I don't know how to proceed as the inverse as well as the matrix form seems complicated to approach directly.

Any ideas??

Answer 1

Using the singular value decomposition, $A$ can be expressed as $$ A=UDV $$ where $U$ and $V$ are unitary and $D$ is a diagonal $m\times n$ matrix. Hence $$ AA^H=UDVV^HD^HU^H=UDD^HU^H $$ and $$ \left[\frac{1}{n}AA^H-zI\right]^{-1}{=\left[\frac{1}{n}UDD^HU^H-zI\right]^{-1} \\= \left[\frac{1}{n}UDD^HU^H-zUU^H\right]^{-1} \\= \left[U(\frac{1}{n}DD^H-zI)U^H\right]^{-1} \\= U\left[\frac{1}{n}DD^H-zI\right]^{-1}U^H }. $$Now the spectral norm should be easy to obtained in this form as the unitary matrices $U$ and $U^H$ are irrelevant.