## 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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