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I have taken this question from Zhang, Fuzhen. Linear Algebra (Johns Hopkins Studies in the Mathematical Sciences)

Assuming that all matrix inverses involved below exist, show that

$(A + iB)^{-1} = B^{-1} A(A + AB^{-1} A)^{-1} − i(B + AB^{-1} A)^{-1}.$

I can simplify the right hand side to obtain $(A + B)^{-1} − i(B + AB^{-1} A)^{-1}$. The hint in the answer key says multiply the right hand side by $(A + iB)$ to verify. But at this point everything is an inverse of the sum. Is there a way to move forward other than using a formula like in this question: Inverse of the sum of matrices?.

Emin Ozkan
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2 Answers2

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There seems to be a typo. The correct formula should be $$ (A+iB)^{-1}=(B^{-1}A-iI)(B+AB^{-1}A)^{-1}. $$ To verify this, multiply the candidate inverse from the left with $A+iB$ to obtain $$ (A+iB)(B^{-1}A-iI)(B+AB^{-1}A)^{-1} =(AB^{-1}A-iA+iA+B)(B+AB^{-1}A)^{-1}=I. $$

J.G.
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daw
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$ \def\B{B^{-1}} \def\C#1{{\mathbb C}^{#1\times #1}} \def\R#1{{\mathbb R}^{#1\times #1}} \def\LR#1{\left(#1\right)} \def\qiq{\quad\implies\quad} \def\qif{\quad\iff\quad} \def\m#1{\left[\begin{array}{r|r}#1\end{array}\right]} \def\CtoR#1#2{\m{#1 & -#2 \\ \hline #2 & #1}} \def\c#1{\color{red}{#1}} $It is often useful to convert a $\:\C{n}$ matrix into its $\,\R{2n}$ representation
$$\eqalign{ A+iB \qif \CtoR AB \\ }$$ Now the problem is to find two real matrices $\LR{E,F}$ which satisfy the identity $$\eqalign{ \m{I&0\\\hline 0&I} = \CtoR AB\bullet\CtoR EF \equiv \m{{AE-BF}&-AF-BE\\\hline {AF+BE}&{AE-BF}} \\ }$$ Solve for the off-diagonal component (assuming $\B$ exists) $$\eqalign{ 0 = AF+BE \qiq E = -\B AF \\ }$$ Then the diagonal component $$\eqalign{ BF &= \LR{A\c{E}-I} \;=\; -A\c{\B AF} - I \\ I &= -\LR{B+A\B A}F \\ F &= -\LR{B+A\B A}^{-1} \\ }$$ Go back and eliminate $F$ from the first equation $$\eqalign{ E &= -\B A\c{F} \;=\; \c{+}\B A\c{\LR{B+A\B A}^{-1}} \\ }$$ Therefore $$\eqalign{ \LR{A+iB}^{-1} &= E \;+\; iF \\ &= \B A\LR{B+A\B A}^{-1} \;\;-\;\; i\LR{B+A\B A}^{-1} \\ }$$ So there was definitely a typo in your textbook.

greg
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