Diagonal bases in transformations

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...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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