# Charecteristic polynomial

1. Feb 25, 2010

### talolard

1. The problem statement, all variables and given/known data
Let A and B be to nXn matrices and A is invertible. Prove: $$P_{AB}(x)=P_{BA}(x)$$

3. The attempt at a solution
Since A is invertible we have det(A)=0. $$det(AB)= det(A)det(B)=0det(B)=0 ->P_{AB}(x)=P_{BA}(x)=0$$

Is that correct?

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2. Feb 25, 2010

### Dick

netheril96, you are supposed to give hints and help. Not work out people's problems for them. Read the forum guidelines.

3. Feb 25, 2010

### netheril96

But this problem is so easy that the only hint I can give is the direct answer

4. Feb 25, 2010

### Dick

Talolard, the determinants you've written down are all just numbers. They don't have anything to do with the characteristic polynomial. What's the definition of a characteristic polynomial in terms of a determinant? Besides if A is invertible it's determinant is NOT 0.

5. Feb 25, 2010

### Dick

It's not 'so easy' for a lot of people. If the only hint you can think of is to give the solution, don't. Find another thread to work on. Could you remove the post with your proof in it please?

6. Feb 25, 2010

### netheril96

Deleted

7. Feb 25, 2010

### Dick

Thanks!

8. Feb 25, 2010

### VeeEight

You may wish to consider similar matrices here