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Let $A$ and $B$ be linear operators on a finite dimensional vector space $V$ over $\mathbb{R}$ such that $AB=(AB)^2$. If $BA$ is invertible then which of the following is true:

(a) $BA = AB$ on $V$
(b) $\operatorname{tr}(A)$ is non zero
(c) $0$ is an eigenvalue of $B$
(d) $1$ is an eigenvalue of $A$

I found that (b) is true but how to show that all other are false?

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  • $\begingroup$ Look for counterexamples? $\endgroup$ – Steven Gubkin Nov 5 '14 at 16:52
  • $\begingroup$ I think (a) is true... $\endgroup$ – Amitai Yuval Nov 5 '14 at 16:56
  • $\begingroup$ How? will you tell? $\endgroup$ – user180150 Nov 5 '14 at 16:57
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If $BA$ is invertible then so are $A,B$. Since $AB=ABAB$, multiplying on the left by $A^{-1}$ and on the right by $B^{-1}$ gives $I = BA$.

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  • $\begingroup$ You can multiply if you know that inverse of A exist. Am I right? $\endgroup$ – user180150 Nov 5 '14 at 17:09
  • $\begingroup$ Yes. You know the inverse exists because $\operatorname{rk} BA \le \min(\operatorname{rk} B,\operatorname{rk} A)$. $\endgroup$ – copper.hat Nov 5 '14 at 17:12
  • $\begingroup$ The point is, to look for counterexamples, just consider any invertible $B$ and then $A$ is automatically given by $B^{-1}$. Diagonal matrices are a useful supply of counterexamples as @Omnomnomnom has noted in the answer below. $\endgroup$ – copper.hat Nov 5 '14 at 17:14
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Hint: If $BA$ is invertible, then $A,B,$ and $AB$ must all be invertible as well.

In fact, of all these statements, only (a) is necessarily true.

Consider the matrices $$ A = \pmatrix{2&0\\0&-2}, \quad B = \pmatrix{1/2 & 0\\0&-1/2} $$ This counterexample shows that every statement except (a) is false.


Hint: as the other answer notes, we can deduce that $AB=I$. Use this to deduce that $AB=BA$. We assume here that both $A$ and $B$ are operators from $V$ to $V$.

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  • $\begingroup$ wonderful!!!. If AB is invertible then how B is invertible? $\endgroup$ – user180150 Nov 5 '14 at 17:04
  • $\begingroup$ Note that $\det(AB) = \det(A) \det(B)$ $\endgroup$ – Omnomnomnom Nov 5 '14 at 17:05
  • $\begingroup$ this shows what sir? $\endgroup$ – user180150 Nov 8 '14 at 15:01
  • $\begingroup$ Note that a matrix is invertible if and only if its determinant is non-zero $\endgroup$ – Omnomnomnom Nov 8 '14 at 15:02

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