# Thread: P-1 A P formula question

1. ## P-1 A P formula question

A , B are 2x2 matrices in which
det A=det B=tr(A)=tr(B)=7
tr- is trace ,the sum of the diagonal

could we link A and B
by
B=P-1AP

2. ## re: P-1 A P formula question

Originally Posted by transgalactic
A , B are 2x2 matrices in which det A=det B=tr(A)=tr(B)=7 tr- is trace ,the sum of the diagonal could we link A and B by B=P-1AP
Hint: The characteristic polynomials of $\displaystyle A$ and $\displaystyle B$ are $\displaystyle \chi(\lambda)=\lambda^2-(\textrm{tr}A)\lambda+\det A=\lambda^2-7\lambda+7$ .

3. ## re: P-1 A P formula question

how you got this polinomial
in ordrer to get The characteristic polynomials of A is |A-KI|
i dont have the matrices of A

4. ## re: P-1 A P formula question

Originally Posted by transgalactic
how you got this polinomial
in ordrer to get The characteristic polynomials of A is |A-KI|
i dont have the matrices of A
He showed you how he got the polynomial. The trace and determinant of a matrix show up as particular coefficients of the characteristic polynomial of that matrix.

5. ## re: P-1 A P formula question

if we have a 2x2 matrix: A =

[a b]
[c d]

then det(A - xI) = (a-x)(d-x) - bc = ad - (a+d)x + x^2 - bc = (ad - bc) - (a+d)x + x^2 = det(A) - tr(A)x + x^2.

6. ## re: P-1 A P formula question

ok i have two eigen values in 2x2 matric
thus making A diagonisable

but it doesnt say that the digonolised form of A is B

?

7. ## re: P-1 A P formula question

Originally Posted by transgalactic
ok i have two eigen values in 2x2 matric
thus making A diagonisable

but it doesnt say that the digonolised form of A is B

?
Well, by the same exact reasoning, wouldn't $\displaystyle B$ also be diagonalizable? And wouldn't $\displaystyle A$ and $\displaystyle B$ have the same diagonalized forms, since they have the same eigenvalues?

8. ## re: P-1 A P formula question

Originally Posted by topspin1617
Well, by the same exact reasoning, wouldn't $\displaystyle B$ also be diagonalizable? And wouldn't $\displaystyle A$ and $\displaystyle B$ have the same diagonalized forms, since they have the same eigenvalues?

In general no, by no means. BUT in the very particular case of 2x2 matrices yes.

Tonio

9. ## re: P-1 A P formula question

Originally Posted by tonio
In general no, by no means. BUT in the very particular case of 2x2 matrices yes.

Tonio
Wait... what are we saying "no" to in general?

I just meant that here, the matrices both have the same set of 2 distinct eigenvalues, so that both will be similar to $\displaystyle \mathrm{diag}(\lambda_1,\lambda_2})$.

10. ## re: P-1 A P formula question

i wasnt asked to compare the digonolised forms.

i was asked to show that there is P in which B=P^-1 A P

11. ## re: P-1 A P formula question

Similarity of matrices is an equivalence relation so, $\displaystyle A\sim D$ and $\displaystyle B\sim D$ implies $\displaystyle A\sim B$ i.e. $\displaystyle P^{-1}AP=B$ for some $\displaystyle P$ .

12. ## re: P-1 A P formula question

Yeah... it's kind of a basic fact that can be proved with arithmetic...

$\displaystyle P^{-1}AP=Q^{-1}BQ\Rightarrow QP^{-1}APQ^{-1}=B\Rightarrow (PQ^{-1})^{-1}A(PQ^{-1})=B$.

13. ## re: P-1 A P formula question

Originally Posted by topspin1617
Wait... what are we saying "no" to in general?

I just meant that here, the matrices both have the same set of 2 distinct eigenvalues, so that both will be similar to $\displaystyle \mathrm{diag}(\lambda_1,\lambda_2})$.

This is not what I understood: you said "And wouldn't and have the same diagonalized forms, since they have the same eigenvalues?", and

to this ia answered "no, in the general case, but yes in the case of dimension 2"...that's all.

Tonio

14. ## re: P-1 A P formula question

Originally Posted by topspin1617
Yeah... it's kind of a basic fact that can be proved with arithmetic...

$\displaystyle P^{-1}AP=Q^{-1}BQ\Rightarrow QP^{-1}APQ^{-1}=B\Rightarrow (PQ^{-1})^{-1}A(PQ^{-1})=B$.
$\displaystyle PQ^-1$ have to be invertible
does multiplication of two invertables is invertible too?

15. ## re: P-1 A P formula question

yes, if A is invertible, and B is invertible, then AB is invertible.

why? since we can form the matrices $\displaystyle A^{-1}$ and $\displaystyle B^{-1}$, the product $\displaystyle B^{-1}A^{-1}$ exists.

but $\displaystyle (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} =AIA^{-1} = AA^{-1} = I$,

and $\displaystyle (B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$.