Linear Algebra and Its Applications, Exercise 1.6.23

Exercise 1.6.23. Assume that A and B are square matrices, and that I - BA is invertible. Show that I - AB is invertible as well. (Use the fact that B(I - AB) = (I - BA)B.)

Answer: First, since A and B are square matrices we know that both of the product matrices AB and BA exist and have the same number of rows and columns. We then have

B(I - AB) = B - BAB = (I - BA)B

Since we are assuming that the inverse of I - BA exists, we have

(I - BA)^{-1} B (I - AB) = (I - BA)^{-1} (I - BA) B = IB = B

Multiplying both sides of the resulting equation on the left by A and then adding I - AB to both sides, we have

A(I - BA)^{-1} B (I - AB) = AB

\rightarrow A(I - BA)^{-1} B (I - AB) + (I - AB) = AB + (I - AB)

\rightarrow [A(I - BA)^{-1} B + I] (I - AB) = AB - AB + I = I

So A(I - BA)^{-1} B + I is a left inverse for I - AB. We then multiply I - AB by A(I - BA)^{-1} B + I on the right:

(I - AB)[A(I - BA)^{-1} B + I]

= (I - AB)A(I - BA)^{-1} B + (I - AB)I

= (A - ABA)(I - BA)^{-1} B + (I - AB)

= A(I - BA)(I - BA)^{-1} B + (I - AB)

= AIB + (I - AB) = AB + I - AB = I

So A(I - BA)^{-1} B + I is also a right inverse for I - AB. Since A(I - BA)^{-1} B + I is both a left inverse and right inverse for I - AB we conclude that I - AB is invertible (with A(I - BA)^{-1} B + I as its inverse).

We have thus showed that if I - BA is invertible then I - AB is also invertible.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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