# $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.

The problem is from Algebra by Artin, Chapter 1, Miscellaneous Problems, Exercise 8. I have been trying for a long time now to solve it but I am unsuccessful.

Let $A,B$ be $m \times n$ and $n \times m$ matrices. Prove that $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.

Please provide with only a hint.

## 4 Answers

Another method:

Let $C= {(I_m - AB)}^{-1}$. The matrix $BCA$ is $n\times n$.

$(I_n - BA)(BCA) = BCA - BABCA$

$= B(C - ABC)A$ $= B[(I_m - AB)C]A$ $= B(I_m)A\$ (by definition of $C$)

$= BA$

Hence,

$(I_n-BA)(BCA + I_n) = (I_n - BA) (BCA) + (I_n -BA) = BA + (I_n-BA) = I_n$

So, we get that the inverse of $I_n - BA$ is $I_n +BCA$.

• This is how I did the problem the first time, but then I forgot how I did it and was stuck. :P Thanks :-) – user14082 Nov 15 '12 at 7:41

There's no real need to actually invoke Sylvester's determinant theorem (although that certainly is much faster).

First show that the (non-zero) eigenvalues of $AB$ and the eigenvalues of $BA$ coincide. If you take the determinant of $I_m - AB$, then you have the characteristic polynomial of $AB$ evaluated at $\lambda = 1$. It follows that the determinant is zero if and only if $1$ is an eigenvalue of $AB$ if and only if $1$ is an eigenvalue of $BA$ if and only if $\det(I_n - BA) = 0$.

Note that a slight adaptation of this argument also provides a proof of Sylvester's determinant theorem different from the one given on Wikipedia.

• Thank you for another nice answer. – user14082 Nov 15 '12 at 6:53
• I like this proof so much better than the others. – Sahiba Arora Nov 8 '15 at 12:59

Hint $\$ It's a consequence of Sylvester's determinant identity $\rm\:det(1 + AB) = det(1+BA),\:$ which has a very simple universal proof:  over the polynomial ring $\rm\ \mathbb Z[A_{\,i\,j},B_{\,i\,j}\,]\$ take the determinant of $\rm\, (1+AB)\, A = A\, (1+BA)\$ then cancel $\rm\, det(A)\$ (valid since the ring is a domain).  Extend to non-square matrices by padding appropriately with $0$'s and $1$'s to get square matrices. Note that the proof is purely algebraic - it does not require any topological notions (e.g. density).

Alternatively $\$ Proceed by way of Schur decomposition, namely

$$\rm\left[ \begin{array}{ccc} 1 & \rm A \\ \rm B & 1 \end{array} \right]\ =\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm 0 & \rm 1-BA \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]$$

$$\rm\phantom{\left[ \begin{array}{ccc} 1 & \rm B \\ \rm A & 1 \end{array} \right]}\ =\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]\ \left[ \begin{array}{ccc} \rm 1-AB & \rm 0 \\ \rm 0 & \rm 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]$$

See my posts in this sci.math thread on 09 Nov 2007 for further discussion.

We know that a square matrix $M$ is invertible if and only if $$\{\vec x; \vec x M = \vec 0\}=\{\vec 0\}.$$ (The corresponding homogeneous linear system has only trivial solution. The corresponding linear map has trivial kernel.)

So the claim $$I_m-AB\text{ is invertible }\qquad\implies\qquad I_n-BA\text{ is invertible}$$ is equivalent to $$\{\vec x\in F^m; \vec x(I-AB)=\vec 0\}=\{\vec0\} \qquad\implies\qquad \{\vec y\in F^n; \vec y(I-BA)=\vec 0\}=\{\vec0\}.$$ (Where $F$ denotes the base field.)

Proof. So let us assume that $I-AB$ is invertible, i.e., that $\vec x=\vec 0$ is the only vector such that $\vec x(I-AB)=\vec 0$.

Let $\vec y(I-BA)=\vec 0$.

Then we get $$\vec y=\vec yBA\tag{1}$$ and, consequently $\vec y B = \vec y BAB$. This is equivalent to $$\vec yB(I-AB)= \vec 0.\tag{2}$$ This means that $\vec yB=\vec 0$, since $I-AB$ is invertible.

From (1) we now get that $\vec y=\vec yBA=\vec 0A=\vec 0$.
$\square$