[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.1

For fixed basis of in $\scrH$ and $\scrK$, the matrix $A^*$ is the conjugate transpose of the matrix of $A$.

Solution.  $$\beex \bea (A^*)_{ij}&=e_i^*A^*f_j\\ &=\sef{e_i,A^*f_j}_\scrH\\ &=\sef{Ae_i,f_j}_\scrK\\ &=\overline{\sef{f_j,Ae_i}}\\ &=\overline{a}_{ji}. \eea \eeex$$

时间： 2024-10-11 14:21:11

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