I am trying to solve Hughston Tod's Problem 2.10)

The Transpose of a matrix $A_{ij}$ is the matrix $\tilde{A}_{ij}=A_{ji}$. A matrix is called symmetric if $A_{ij}=\tilde{A}_{ij}$, and antisymmetric if $A_{ij}=-\tilde{A}_{ij}$. A matrix is called orthogonal if $A_{ij}\tilde{A}_{jk}=\delta_{ik}$. Suppose $P_{ij}$ is antisymmetric, and that $\delta_{ij}+P_{ij}$ has an inverse $Q_{ij}$ Show that $A_{ik}=(\delta_{ij}-P_{ij})Q_{jk}$ is orthogonal.

I went about this problem in the following way:

If $$A_{ik}=(\delta_{ij}-P_{ij})Q_{jk}$$ then $$\tilde{A}_{ij}=(\delta_{ik}+P_{ik})\tilde{Q}_{kj}$$ Then multiplying $$A_{ik}\tilde{A}_{ij}=(\delta_{in}-P_{in})\underbrace{Q_{nk}(\delta_{km}+P_{km})}_{=\delta_{nm}}\tilde{Q}_{mj}=\delta_{ij}$$ The last equality can be verified by taking the Transpose of the term in the braces.

Did I do everything correct?

  • $\begingroup$ In more traditional notation, you're being asked to show that the Cayley transform of a skew symmetric matrix, $(I-P)(I+P)^{-1}$, is orthogonal. $\endgroup$ – Omnomnomnom Dec 14 '16 at 3:07
  • $\begingroup$ Proofs in the usual (non Einstein-summation) notation are given here. Whichever notation you use, it's important to keep matrices in the correct order and justify any "switching around". $\endgroup$ – Omnomnomnom Dec 14 '16 at 3:10
  • $\begingroup$ I don't quite see how the problem I defined is related to the Cayley Transform of a skew symmetric matrix. $\endgroup$ – onephys Dec 14 '16 at 7:42
  • $\begingroup$ do you not see that $Q=(I+P)^{-1}$? $\endgroup$ – Omnomnomnom Dec 14 '16 at 9:35
  • $\begingroup$ No I don't quite see that. $\endgroup$ – onephys Dec 14 '16 at 13:02

You took the transpose incorrectly. You should find $$ \tilde A_{ij} = (\delta_{jk} - P_{jk})Q_{ki} = Q_{ki}(\delta_{jk} - P_{jk}) = \tilde Q_{ik} (\delta_{kj} + P_{kj}) $$

  • $\begingroup$ how does the rest then work out? cause then you have the Q's multiplied and they don't yield the identity? $\endgroup$ – onephys Dec 14 '16 at 7:34

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