# antisymmetric matrix show that orthogonal

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.

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?

• 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. – Omnomnomnom Dec 14 '16 at 3:07
• 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". – Omnomnomnom Dec 14 '16 at 3:10
• I don't quite see how the problem I defined is related to the Cayley Transform of a skew symmetric matrix. – onephys Dec 14 '16 at 7:42
• do you not see that $Q=(I+P)^{-1}$? – Omnomnomnom Dec 14 '16 at 9:35
• No I don't quite see that. – onephys Dec 14 '16 at 13:02

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