# P-1 A P formula question

• June 22nd 2011, 02:23 PM
transgalactic
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
• June 22nd 2011, 02:34 PM
FernandoRevilla
re: P-1 A P formula question
Quote:

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 $A$ and $B$ are $\chi(\lambda)=\lambda^2-(\textrm{tr}A)\lambda+\det A=\lambda^2-7\lambda+7$ .
• June 22nd 2011, 02:43 PM
transgalactic
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
• June 22nd 2011, 02:44 PM
topspin1617
re: P-1 A P formula question
Quote:

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.
• June 22nd 2011, 02:55 PM
Deveno
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.
• June 22nd 2011, 03:10 PM
transgalactic
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

?
• June 22nd 2011, 06:09 PM
topspin1617
re: P-1 A P formula question
Quote:

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 $B$ also be diagonalizable? And wouldn't $A$ and $B$ have the same diagonalized forms, since they have the same eigenvalues?
• June 22nd 2011, 07:24 PM
tonio
re: P-1 A P formula question
Quote:

Originally Posted by topspin1617
Well, by the same exact reasoning, wouldn't $B$ also be diagonalizable? And wouldn't $A$ and $B$ 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
• June 22nd 2011, 07:51 PM
topspin1617
re: P-1 A P formula question
Quote:

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 $\mathrm{diag}(\lambda_1,\lambda_2})$.
• June 22nd 2011, 11:35 PM
transgalactic
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
• June 23rd 2011, 12:20 AM
FernandoRevilla
re: P-1 A P formula question
Similarity of matrices is an equivalence relation so, $A\sim D$ and $B\sim D$ implies $A\sim B$ i.e. $P^{-1}AP=B$ for some $P$ .
• June 23rd 2011, 01:31 AM
topspin1617
re: P-1 A P formula question
Yeah... it's kind of a basic fact that can be proved with arithmetic...

$P^{-1}AP=Q^{-1}BQ\Rightarrow QP^{-1}APQ^{-1}=B\Rightarrow (PQ^{-1})^{-1}A(PQ^{-1})=B$.
• June 23rd 2011, 02:39 AM
tonio
re: P-1 A P formula question
Quote:

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 $\mathrm{diag}(\lambda_1,\lambda_2})$.

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
• June 25th 2011, 12:02 PM
transgalactic
re: P-1 A P formula question
Quote:

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

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

$PQ^-1$ have to be invertible
does multiplication of two invertables is invertible too?
• June 25th 2011, 12:35 PM
Deveno
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 $A^{-1}$ and $B^{-1}$, the product $B^{-1}A^{-1}$ exists.

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

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