# Eigen value and transpose of the $Matrix$

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.

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

• @ Jean : what can you say about the polylomial $x^2 -x$ which divides the characterstic polynomian of $AB$ – user120386 Apr 24 '15 at 11:58
• @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. – Stop hurting Monica Apr 24 '15 at 12:00
• @ Jean : thanks for your prompt reply. – user120386 Apr 24 '15 at 12:02
• @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 $Q$. – 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 $AB = I.$