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Here $A^{T}$ is transpose matrix of $A$, $I$ is identity matrix.

Also prove

1) $(I-A)(I+A) = (I+A)(I-A)$

2) $(I-iA)(I+iA) = (I+iA)(I-iA)$

3) $(I-iA)^{-1} (I+iA)^{-1} = (I+iA)^{-1}(I-iA)^{-1}$

These properties are being used in a question I'm trying to solve, please give me a hint to prove.

abhinav12369
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1 Answers1

0

For all matrices $M$, \begin{align} (I-M)(I+M)&=\\ &=(I-M)I+(I-M)M\\ &=I\cdot I-M\cdot I+I\cdot M-M\cdot M\\ &=I-M+M-M^2\\ &=I-M^2\\ &=I\cdot I+M\cdot I-I\cdot M-M\cdot M\\ &=(I+M)I-(I+M)A\\ &=(I+M)(I-M) \end{align} Plugging $M=A$ gives $(1)$.

Plugging $M=iA$ gives $(2)$.

As for any invertivble $M,N$ $$MNN^{-1}M^{-1}=I=N^{-1}M^{-1}MN$$ we have $$(MN)^{-1}=N^{-1}M^{-1}$$

Then, using $(2)$,

$$(I-iA)^{-1} (I+iA)^{-1}=\left((I-iA)\cdot(I+iA)\right)^{-1} = \left((I+iA)\cdot(I-iA)\right)^{-1} = (I+iA)^{-1}(I-iA)^{-1}$$

Emre
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