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Exercise 2.5.44

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Inverting the identity $A (I + BA) = (I + AB) A$ gives ${(I + BA)}^{- 1} A^{- 1} = A^{- 1} (I +$ $AB)^{- 1}$ . So $I + BA$ and $I + AB$ are both invertible or both singular when $A$ is invertible. (This remains true also when $A$ is singular: Chapter 6 will show that $AB$ and $BA$ have the same nonzero eigenvalues, and we are looking here at the eigenvalue $- 1$ .)

mathchakra

2021-12-19 17:45

Exercise 2.5.44

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