antisymmetric matrix show that orthogonal

Question

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?

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. — Ben Grossmann, Dec 14 '16 at 03:07
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". — Ben Grossmann, Dec 14 '16 at 03:10
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 07:42
Alright. First, note that the matrix whose entries are $\delta_{ij}$ is the identity matrix, commonly denoted by $I$. So, $\delta_{ij} + P_{ij}$ is the same as $I + P$. So, the inverse of that matrix (which is our $Q$) is $(I+P)^{-1}$. Does that make sense now? — Ben Grossmann, Dec 14 '16 at 13:04

Ben Grossmann · Accepted Answer · 2016-12-14T03:27:20.343

1

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

edited Dec 14 '16 at 03:27

answered Dec 14 '16 at 02:50

Ben Grossmann

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how does the rest then work out? cause then you have the Q's multiplied and they don't yield the identity? – onephys Dec 14 '16 at 07:34

antisymmetric matrix show that orthogonal

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