how to find eigenvalues/eigenvectorsby mind0nmath Tags: None 

#1
Mar408, 04:02 PM

P: 19

How do i find the eigenvalues and eigenvectors for the linear operator T defined as
T(w,z) = (z,w)?? 



#2
Mar408, 04:19 PM

P: 270





#3
Mar508, 11:31 AM

P: 19

how about for something like: T(x_1,x_2,...,x_n) = (x_1+x_2+...+x_n, x_1+x_2+...+x_n, ..., x_1+x_2+...+x_n). The matrix with respect to standard basis would have 1's everywhere? any clues to finding the eigenvalues/vectors?




#4
Mar508, 02:32 PM

P: 341

how to find eigenvalues/eigenvectors
Try Matlab command >>[V,D] = eig(ones(n))




#5
Mar508, 02:52 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

??? T operates on a pair of numbers and gives a pair of numbers as the result. Written as a matrix, it would be 2 by 2 matrix certainly not as complicated as you have! You are not still referring to the first problem are you?
By definition, T(w,z)= (z, w) so T(1, 0)= (0, 1) and T(0, 1)= (1, 0). [tex]\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{c} 1 \\ 0\end{array}\right)= \left(\begin{array}{c} 0 & 1\end{array}\right)[/tex] What are a and c? Do the same with (0, 1) being taken to (1, 0) to determine b and d. A good way of determining the matrix representing a linear operator in a given basis is to apply it to each of the basis vectors in turn. The result will be a column of the matrix. Of course, you don't have to write it as a matrix to find eigenvalues in fact, I think too many students get the idea that Linear Algebra is only about matrices. Saying that [itex]\lambda[/itex] is an eigenvalue for linear transformation T means that there exist some (x, y), not both 0, such that [itex]T(x,y)= \lambda(x, y)= (\lambda x, \lambda y)[/itex]. Since T(x,y)= (y, x), that says that [itex](y, x)= (\lambda x, \lambda y)[/itex] so you have two equations: [itex]y= \lambda x[/itex] and [itex]x= \lambda y[/itex]. Obviously, x= y= 0 would satisfy those equations for any [itex]\lambda[/itex]. For what values of [itex]\lambda[/itex] would that have nonzero solutions? If you replace the "x" in the first equation by [itex]\lambda y[/itex] from the second equation, you have [itex]y= \lambda(\lambda y)= \lambda^2 y[/itex]. If y is not 0, you can divide both sides by y to get [itex]\lambda^2= 1[/itex]. 



#6
Mar508, 03:01 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Okay, what are the eigenvalues of [tex]\left(\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right)[/tex]? (Hint: if a matrix has two rows (or two columns) the same it has determinant 0. And if it has deteminant 0, it has 0 as an eigenvalue.) The eigenvalues must satisfy [tex]\left\begin{array}{cc} 1\lambda & 1 \\ 1 & 1 \lambda\end{array}\right= 0[/tex] What equation does that give you? What are the eigenvalues? 


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