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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.

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$(iii)\, AB$ 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.

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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, $\det(B)=0$.

Hence $\det(AB)=\det(BA)=0$

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 $Bv$.

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