Let A∈Mn(R)"> A ∈ M n ( R ) A∈Mn(R) such that I+A"> I + A I+A is invertible. Suppose that

Let $A \in M n (R) "> A \in M_{n} (R)$ such that $I + A "> I + A$ is invertible. Suppose that

$B = (I - A) (I + A) - 1 "> B = (I - A) (I + A)^{- 1}$

(a) Show that $B = (I + A) - 1 (I - A) "> B = (I + A)^{- 1} (I - A)$

(b) Show that $I + B "> I + B$ is invertible and express $A "> A$ in terms of $B "> B$ .

Expert Solution

Solution

(a) Show that $B = (I + A) - 1 (I - A) "> B = (I + A)^{- 1} (I - A)$ We have $(I + A) (I - A) = I - A + A - A 2 (I - A) (I + A) = I + A - A - A 2 (I + A) (I - A) = (I - A) (I + A) (I - A) = (I + A) - 1 (I - A) "> \begin{matrix} (I + A) (I - A) & = I - A + A - A^{2} (I - A) (I + A) & = I + A - A - A^{2} (I + A) (I - A) & = (I - A) (I + A) (I - A) & = (I + A)^{- 1} (I - A) \end{matrix}$ Muhiply from the right side by $(I + A) - 1 "> (I + A)^{- 1}$

then we get $(I - A) (I + A) - 1 = (I + A) - 1 (I - A) ">$ $(I - A) (I + A) - 1 = (I + A) - 1 (I - A) "> (I - A) (I + A)^{- 1} = (I + A)^{- 1} (I - A)$

Therefore, $B = (I + A) - 1 (I - A) "> B = (I + A)^{- 1} (I - A)$

(b) Show that $I + B "> I + B$ is invervible and express $A "> A$ in terms of $B "> B$ We have $(I + A) (I + B) = (I + A) + (I + A) B = I + A + I - A = 2 I "> \begin{matrix} (I + A) (I + B) & = (I + A) + (I + A) B = I + A + I - A = 2 I \end{matrix}$ Therefore, $(I + B) "> (I + B)$ is invertible and its inverse is $12 (I + A) $ T h e n "> \frac{1}{2} (I + A) ,$ $12 (I + A) $ T h e n "> T h e n$ $(I - B) (I + B) - 1 = 12 (I - B) (I + A) = 12 [I + A - B (I + A)] = 12 (I + A - I + A) (I - B) (I + B) - 1 = A "> \begin{matrix} (I - B) (I + B)^{- 1} & = \frac{1}{2} (I - B) (I + A) = \frac{1}{2} [I + A - B (I + A)] = \frac{1}{2} (I + A - I + A) (I - B) (I + B)^{- 1} & = A \end{matrix}$ $

SUN BUNRA answered 2 years ago

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Question

Let A∈Mn(R)"> A ∈ M n ( R ) A∈Mn(R) such that I+A"> I + A I+A is invertible. Suppose that

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Expert Solution

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