Let $A$ and B be $n\times n$ matrices over a field $F$. Then prove or disprove
$(i)\, AB$ and $BA$ have same characteristic values.
$(ii)\, AB$ and $BA $ have same characteristic polynomial.
$(iii)\, AB$ and $BA $ have same minimal polynomial.
Proof of $(i)$
Let $c$ be an eigen value of AB and v be corresponding eigen vector. Then
AB(v)=cv, B(ABv)=B(cv), BA(Bv)=c(Bv),
this implies $Bv$ is an eigenvector of $BA$ corr. to eigen value $c$.
But this statement holds if $Bv$ is nonzero.
What if $Bv=0$?
$(ii)$ If $A$ is invertible then ($A$ inverse)$(AB)A= BA$, i.e. $AB $ and $BA$ are similar and hence have same charateristic polynomial.
Similar proof when $B$ is invertible.
But what if both $A$ and $B$ are not invertible?
I have no idea about $(iii)$.
Please help me to complete these proofs.
First note that there is an equivalent definition for characteristic value – a characteristic value of $A$ is a scalar $c$ such that the matrix $(A-cI)$ is NOT invertible.
For (i) – if $0$ is a characteristic value of $AB$ then $AB$ is not invertible $\Rightarrow BA$ is not invertible and hence $0$ is a characteristic value of $BA$. Let us assume that $c\neq 0$ is a characteristic value of $AB$. Then as you have shown $c$ is a characteristic value of $BA$ as long as $Bv\neq0$. Suppose $Bv=0\Rightarrow A(Bv)=0\Rightarrow cv=0$. Since $c\neq0$ we have $v=0$ which is a contradiction.
For (ii) and (iii) see Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?
ANOTHER PROOF OF AB AND BA HAVE SAME CHARACTERISTIC VALUES.
(Using the result that if (I-AB) is invertible then (I-BA) is invertible.)
We have to show that if x is a characteristic value for AB then x is a
characteristic value for BA (and conversely). This is equivalent to the statement,
if x is not a characteristic value for BA then it is not a characteristic value for
AB. We will prove this last statement.
Suppose that x is not a characteristic value for BA, this means that det(xI−BA) is non zero.
There are two cases:
Case 1: x = 0. In this case det(−BA) is non-zero. But det(−BA) = (−1)^n det(B) det(A) =
(−1)^n det(A) det(B) = det(−AB) = det(xI − AB). Therefore det(xI − AB)is non-zero.
Case 2: x is non-zero.
In this case xI−BA = x(I−1/x BA) and det(x(I−1/x BA)) = x^n det(I−1/x BA)is non-zero. Therefore I −1/x BA is invertible, but this implies
that I − A (1/x)B = I − 1/x AB is invertible, therefore det(I−1/x AB) is non-zero, therefore
x^n det(I −1/x AB) = det(xI − AB) is non-zero.
Proof of AB and BA have same minimal polynomial.
In general, AB and BA do not have same minimal polynomial.
Example: Take A= E2,2 and B= E2,4 , where A and B are 2*2 matrices and Ei,j is the matrix with i,j th entry as 1 and all other entries as 0.
AB and BA have same minimal polynomial when any one of A or B is invertible.
If A is invertible, then (A inverse)(AB)A= BA, i.e. AB and BA are similar.
Now, as similar matrices have same minimal polynomial, so AB and BA have same minimal polynomials.
If $Bv=0$, then as $v$ is non-zero, so $B$ cannot be invertible. So, $\det(B)=0$.
So $AB$ and $BA$ both have eigenvalue $0$.
Not if $c$ is any other non-zero eigenvalue of $AB$ , then $AB(v)=cv$.
Proceeding as you did, if $Bv=0$, then we get $c=0$.
So $Bv$ can’t be zero and hence $c$ is an eigenvalue of $BA$ corresponding to eigenvector $Bv$.