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Let A and B be matrices of order n. Provce that if (I - AB) is invertible, (I - BA) is also invertible and `(I-BA)^(-1) = I + B (I- AB)^(-1)A, ` where I be the dientity matrix of order n. |
Answer» Here, `I -BA = BIB^(-1)-BABB^(-1) = B(I-AB)B^(-1)` ...(i) Hence, `abs(I-BA) = abs(B)abs(I-AB) abs(B^(-1)) = abs(B) abs(I-AB) 1/abs(B)` `= abs(I-AB)` If `abs(I-AB)ne 0, "then" abs(I-BA) ne 0` i.e. if `(I -AB)` is invertible, then `(I - BA)` is also invertible. Now, `(I-BA) [ I + B (I-AB)^(-1)A]` `= (I- BA) + (I-BA) B (I-AB)^(-1)A ` [ using Eq. (i)] `= (I- BA) +B (I-AB) B^(-1) B (I-AB)^(-1)A ` `=(I-BA)B(I-AB)(I-AB^(-1))A` `= (I - BA) + BA= I` Hence, `(I- BA)^(-1)=I + B(I-AB)^(-1) A.` |
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