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1. |
If `A` is a skew symmetric matrix, then `B=(I-A)(I+A)^(-1)` is (where `I` is an identity matrix of same order as of `A`) |
Answer» We have `A=(I+S)(I-S)^(-1)` `:. A^(T)=[(I-S)^(-1)]^(T) (I+S)^(T)=[(I-S)^(T)]^(-1) (I+S)^(T)` But `(I-S)^(T)=I^(T)-S^(T)=I +S" "("as "S^(T)=-S)` and `(I+S)^(T)=I^(T)+S^(T)=I-S` `:. A^(T)=(I+S)^(-1) (I-S)` `:. A^(T)A=(I+S)^(-1) (I-S) (I+S) (I-S)^(-1)` `=(I+S)^(-1) (I+S) (I-S) (I-S)^(-1)` `=I` Thus, a is orthogonal. |
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