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

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- Jun 22nd 2011, 02:23 PMtransgalacticP-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 - Jun 22nd 2011, 02:34 PMFernandoRevillare: P-1 A P formula question
- Jun 22nd 2011, 02:43 PMtransgalacticre: 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 - Jun 22nd 2011, 02:44 PMtopspin1617re: P-1 A P formula question
- Jun 22nd 2011, 02:55 PMDevenore: 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. - Jun 22nd 2011, 03:10 PMtransgalacticre: 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

? - Jun 22nd 2011, 06:09 PMtopspin1617re: P-1 A P formula question
- Jun 22nd 2011, 07:24 PMtoniore: P-1 A P formula question
- Jun 22nd 2011, 07:51 PMtopspin1617re: P-1 A P formula question
- Jun 22nd 2011, 11:35 PMtransgalacticre: 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 - Jun 23rd 2011, 12:20 AMFernandoRevillare: 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 $\displaystyle P$ .

- Jun 23rd 2011, 01:31 AMtopspin1617re: P-1 A P formula question
Yeah... it's kind of a basic fact that can be proved with arithmetic...

$\displaystyle P^{-1}AP=Q^{-1}BQ\Rightarrow QP^{-1}APQ^{-1}=B\Rightarrow (PQ^{-1})^{-1}A(PQ^{-1})=B$. - Jun 23rd 2011, 02:39 AMtoniore: P-1 A P formula question
- Jun 25th 2011, 12:02 PMtransgalacticre: P-1 A P formula question
- Jun 25th 2011, 12:35 PMDevenore: 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 $\displaystyle (B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$.