Diagonal bases in transformations

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Homework Help Overview

The problem involves finding a diagonal basis for a linear operator T defined on R3. The original poster presents the operator's matrix and its characteristic equation, raising concerns about the nature of its roots.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the characteristic equation derived from the matrix of the operator and question the existence of real solutions. There is an exploration of the implications of the matrix being symmetric and the potential for real eigenvalues.

Discussion Status

Some participants have confirmed the characteristic equation and are collaboratively examining the roots. There is a recognition of a potential error in the original poster's calculations, with suggestions to reconsider the roots of the cubic equation.

Contextual Notes

The original poster mentions using Excel to find roots, indicating possible computational errors. The discussion reflects uncertainty about the nature of the solutions to the characteristic equation.

derryck1234
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Homework Statement



Let T: R3 - R3 be the linear operator given by

T = -y + z
-x + z
x + y

Find a basis B' for R3 relative to which the matrix for T is diagonal using the standard basis B for R3.

Homework Equations



[T]B' = P-1[T]BP

The Attempt at a Solution



I find the standard matrix for T to be

0 -1 1
-1 0 1
1 1 0

The characteristic equation of which, I find to be

(lambda)^3 -3(lambda) + 2 = 0

Which has no real solutions? What can I do?

Thanks

Derryck
 
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well the matrix is symmetric so that should ensure real eigenvalues...

from visual inspection it appears 1 is a root
 
Hi Derryck! :smile:

(have a lambda: λ and try using the X2 icon just above the Reply box :wink:)
derryck1234 said:
(lambda)^3 -3(lambda) + 2 = 0

Which has no real solutions?

erm :redface:

how can a cubic equation have no real solutions? :wink:
 
I also got the same characteristic equation as well...
 
Ok thanks guys. See the thing is I just put it into excel to help me find roots. I must have entered the wrong formula though:( It came up with an irrational number? Anyway...I can definitely see that 1 is a root now...thanks...
 

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