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Jul 17, 2023 at 14:59 answer added jlewk timeline score: 0
Jul 17, 2023 at 12:52 answer added loup blanc timeline score: -1
Jul 17, 2023 at 3:36 comment added jlewk I suggest the experiment (from, e.g, matlab) to draw histograms of eigenvalues of $AA^T/n$ and of the matrix $(AA^T/n + I_m)$ of interest. The eigenvalues of $AA^T/n$ would approximately follow the Marcenko-Pastur law, so the two matrices are very far from proportional to identity when m/n is fixed.
Jul 17, 2023 at 3:01 comment added jlewk What does $\frac1n AA^T \to I_m$ mean when $n,m\to+\infty$ (both) with fixed $m/n$?
Jul 15, 2023 at 20:51 comment added Thomas Kojar i suppose one could also try the Portmanteau theorem, but that would require proving continuity of the above inverse procedure.
Jul 15, 2023 at 20:47 comment added Thomas Kojar well since this matrix $M=AA^{H}$ is self-adjoint, you have a eigendecomposition and thus a nice formula for the inverse math.stackexchange.com/questions/298616/what-is-inverse-of-ia \begin{equation} \left( A + I\right)^{-1} =\left( Q \Lambda Q^{-1} + I\right)^{-1} =\left( Q \left(\Lambda + I \right) Q^{-1}\right)^{-1} = Q \left(\Lambda + I \right)^{-1} Q^{-1} \end{equation}
Jul 15, 2023 at 19:19 history asked Math_Y CC BY-SA 4.0