Question

If $A \in \mathrm{M}_n$ is normal and $A=U \Lambda U^$ is a spectral decomposition, show that $A^{\dagger}=U \Lambda^{\dagger} U^$.

   If $A \in \mathrm{M}_n$ is normal and $A=U \Lambda U^*$ is a spectral decomposition, show that $A^{\dagger}=U \Lambda^{\dagger} U^*$.

A Second Course in Linear Algebra

A Second Course in Linear Algebra

Stephan Ramon… 1st Edition

Chapter 15, Problem 22

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Step 1

If $A$ is a normal matrix, then $A^\dagger$ can be computed using the spectral decomposition of $A$. The spectral decomposition of a normal matrix $A$ is given by $A = U \Lambda U^*$, where $U$ is a unitary matrix (i.e., $U^* U = U U^* = I$, the identity matrix) Show more…

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If $A \in \mathrm{M}_n$ is normal and $A=U \Lambda U^*$ is a spectral decomposition, show that $A^{\dagger}=U \Lambda^{\dagger} U^*$.

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