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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- June 22nd 2011, 03:23 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 - June 22nd 2011, 03:34 PMFernandoRevillare: P-1 A P formula question
- June 22nd 2011, 03:43 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 - June 22nd 2011, 03:44 PMtopspin1617re: P-1 A P formula question
- June 22nd 2011, 03:55 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. - June 22nd 2011, 04:10 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

? - June 22nd 2011, 07:09 PMtopspin1617re: P-1 A P formula question
- June 22nd 2011, 08:24 PMtoniore: P-1 A P formula question
- June 22nd 2011, 08:51 PMtopspin1617re: P-1 A P formula question
- June 23rd 2011, 12:35 AMtransgalacticre: 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 - June 23rd 2011, 01:20 AMFernandoRevillare: P-1 A P formula question
Similarity of matrices is an equivalence relation so, and implies i.e. for some .

- June 23rd 2011, 02:31 AMtopspin1617re: P-1 A P formula question
Yeah... it's kind of a basic fact that can be proved with arithmetic...

. - June 23rd 2011, 03:39 AMtoniore: P-1 A P formula question
- June 25th 2011, 01:02 PMtransgalacticre: P-1 A P formula question
- June 25th 2011, 01:35 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 and , the product exists.

but ,

and .