Let $A$ and $B$ are linear operator on a finite dimensional vector space $V$ over $\mathbb R$ such that $(AB)^2 = AB$ and $BA$ is invertible , then which of the following are true ?

  1. $AB = BA$ on $V$.

  2. $ Tr (A)$ is non zero

  3. $0$ is an eigen value of $B$

  4. $1$ is an eigen value of$A$

Since $BA$ is invertible , then $A$ and $B$ are invertible , and eigen value of $AB$ are $0$ and $1$ and $AB$ is dioganlizabe , because minimal polynomial of $AB$ is $x^2 -x = 0 $, and we know that the eigen value of $AB$ and $BA$ are same counted with the multiplecity of non -zeroeigen value and $BA$ is invertible . So the eigen value of $BA$ is $1$ with multiplecity $n$.

I am confused. my concept is contradicting.please give a me way how to solve.

Thank you.

  • $\begingroup$ Which linear operator has eigenvalue $1$ with geometric multiplicity $\dim V$? What is then the relation between $A$ and $B$? $\endgroup$ – levap Apr 24 '15 at 11:49
  • $\begingroup$ @ Levap : $AB$ has eigen value $1$ with multiplicity $dim \ V$ $\endgroup$ – user120386 Apr 24 '15 at 12:05

$A$ and $B$ are square matrices. Since $BA$ is invertible, $A$ and $B$ are both invertible, because $\det(B)\det(A)=\det(BA)\neq 0$.

Thus $ABAB=AB$ implies $AB=I$, and $B$ is the inverse of $A$, thus $AB=BA=I$.

Since $B$ is invertible, it does not have $0$ as an eigenvalue.

You can't say anything about (2) and (4) because all invertible matrices $A$ satisfy your conditions. And of course, not all invertible matrice satisfy (2) or (4), though some do (for example the identity matrix).

A trivial counter example to (2) is a permutation matrix associated to a derangement (thus only $0$s on the diagonal). A trivial counter example to (4) is any multiple $aI$ of the identity matrix with $a\neq 1$.

  • $\begingroup$ @ Jean : what can you say about the polylomial $x^2 -x$ which divides the characterstic polynomian of $AB$ $\endgroup$ – user120386 Apr 24 '15 at 11:58
  • $\begingroup$ @user120386 Wrong, since $AB=I$ and the characteristic polynomial is thus $(x-1)^n$. You have never proved that $x^2-x$ is the minimal polynomial. $\endgroup$ – Stop hurting Monica Apr 24 '15 at 12:00
  • $\begingroup$ @ Jean : thanks for your prompt reply. $\endgroup$ – user120386 Apr 24 '15 at 12:02
  • 1
    $\begingroup$ @user120386 Actually, what is troublesome here: if $P(A)=0$, then it's a multiple of a minimal polynomial (say $Q$), because $\Bbb R[X]$ is a principal ring, and polynomials $P$ such that $P(A)=0$ obviously constitute an ideal. But, given some polynomial $P$ such that $P(A)=0$, you only know that $P$ is a multiple of $Q$. $\endgroup$ – Stop hurting Monica Apr 24 '15 at 12:06

can we do this? it looks like $AB = I.$ here is the reason. we have $$I = (AB)^2 = ABAB = AB$$ multiplying by $A$ on the right and $(BA)^{-1}$ on the left gives us $$ABA(BA)(BA)^{-1}=A(BA)(BA)^{-1}\to ABA = A \to (BA)^2 = (BA) $$ and that $BA$ is invertible implies that only eigenvalues of $BA$ is $1.$

i think $(BA)^2 = BA$ and $1$ is the only eigenvalue must imply that $$BA = I $$ which in turn give $AB = I.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.