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Unformatted text preview: 1 Eigenvalues and eigenvectors  wtf? In general, a linear transformation T maps any line L R n to some other line in R n . Of course, there are a bazillion different lines in R n , so there is no system atic way to understand T by looking at random lines. The key is to try to see if T maps any lines L to itself. Eigenvectors are vectors that are the oneelement bases for such lines (and as such, must be nonzero). If T does in fact map L to itself then it must do so by scalar multiplication; it maps L to itself by the formula T~x = ~x . This scalar is what an eigenvalue is. Notice that different kinds of eigenvalues have different qualitative behavior. If the eigenvalue is 1, then T~x = 1 ~x = ~x , so T acts as the identity along L . If = 0 then T~x = 0 ~x = 0, so T is collapsing L to the origin. When = 1, then T~x = 1 ~x = ~x , so T flips, or reflects, the line. When   < 1 then T is some kind of contraction and when   > 1, T is dialating L . Exercise 1.1. Suppose T is a linear transformation, and is an eigenvalue for T . Show the set of vectors such that T~x = ~x is a subspace. Such a space is called the eigenspace of . An eigenbasis for T is a basis { ~x 1 ,...,~x n } of R n such that each ~x i is an eigen vector for T . Here are the important facts: Many linear transformations dont have an eigenbasis! Linear transformations that have an eigenbasis are easy to understand. They are diagonalizable 2 Change of basis and eigenbasis First we need to see why linear transformations that have an eigenbasis are easy to understand. Let { ~x 1 ,...,~x n } be a basis and let T : R n R n be linear transformation. Then the matrix that represents T in coordinates is [ T ] = [ T~x 1 ] [ T~x n ] Now if is an eigenbasis, then each of the ~x i s is an eigenvector with eigenvalue say i . So [ T~x i ] = [ i ~x i ] = . . . i . . . which is the vector consisting of all zeros except for the i th component which is i . Plugging this in to our formula for 1 [ T ] yields 1 2 . . . . . . . . . n aka the diagonal matrix whose i th diagonal component is i . Exercise 2.1. Suppose T : R 4 R 4 is a linear transformation and = { ~v 1 ,~v 2 ,~v 3 ,~v 4 } is an eigenbasis for T with corresponding eigenvalues 2 , 4 , , and 1 . What is [ T ] ? solution: [ T ] = 2 4 1 . 3 Using an eigenbasis to find matrix represen tations Suppose there are two islands, and . On , the inhabitants keep lush groves of cheese trees and keep griddles well larded and hot for making grilled cheese sandwiches. On however the foolish inhabitants long ago forgot how to care for...
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This note was uploaded on 01/11/2012 for the course MATH 100 taught by Professor Loveys during the Fall '11 term at McGill.
 Fall '11
 Loveys
 Eigenvectors, Vectors

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