A question about a new way to find eigenvectors that i noticed

[transgalactic]

i got this question in which we are given the matrix T
and we need to find the eigenvalues and the independent spaces (i don't know what is independent space) of T^2 +2*T

the problem is that he started to solve the question as i would have solved it
but then he puts a big X on it and does something else
i can't understand it??(and he gets all the point for it)

it looks as if he skips the finding the roots of polinomial step
why?

http://img253.imageshack.us/my.php?image=img86091xg4.jpg

 

[Dick]

The solver appears to have realized that he didn't have to compute the eigenvalues of T^2+2*T since he already knew that the eigenvalues of T were +1 and -1, apparently from a previous problem. This let him immediately conclude the eigenvalues of T^2+2*T are 3 and -1. Once he knew the eigenvalues he substituted them in for lambda and seems to have read off the eigenvectors more or less by inspection. It's not a new way of computing eigenvalues.

 

[HallsofIvy]

If $\lambda$ is an eigenvalue of T, with eigenvector v, then Tv= $\lambda$v. From that, $(T^2+ 2T)v= T(T(v))+ 2T(v)= T(\lambda v)- 2\lambda= \lambda T(v)- 2\lambda= \lambda(\lambda v)- 2\lambda v= (\lambda^2- 2\lambda) v$.

In other words if $\lambda$ is an eigenvalue of T with eigenvector v, then $\lambda^2- 2\lambda$ is an eigenvalue of T²- 2T with eigenvector v.

It is easier to find the eigenvalues of T and then use that formula than to find the eigenvalues of T²- 2T directly.

 

[transgalactic]

ok i understood how you got the formula from the T^2 + 2T epression

what now??
how do i mix up the eigenvalues of T with this formula in order to get the new values
??

 

[HallsofIvy]

ok i understood how you got the formula from the T^2 + 2T epression

what now??
how do i mix up the eigenvalues of T with this formula in order to get the new values
??

What do you mean by "mix up the eigenvalues of T" and what "new values" are you talking about?

If you mean "How do I go from the eigenvalues of T to the eigenvalues of T²- 2T?", that's exactly what I told you before.:

In other words if $\lambda$ is an eigenvalue of T with eigenvector v, then $\lambda^2- 2\lambda$ is an eigenvalue of T²- 2T with eigenvector v.

 

[transgalactic]

correct me if i am wrong

x-eigenvalue of T
y-eigen value of the expression
y(x)=x^2-2*x

so if x=-1 then for that "old" eigen value we get y=3
and we do that process for every eigenvalue