204Lecture102006 - Economics 204 Lecture 10Friday, August...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Economics 204 Lecture 10–Friday, August 11, 2006 Revised 8/11/06–Revisions indicated by ** Section 3.5, Change of Basis, Similarity Let X be a fnite-dimensional vector space with basis V .I F T L ( X, X ) it is customary to use the same basis in the domain and range: De±nition 1 Mtx V ( T ) denotes V,V ( T ) Question: IF W is another basis For X ,howare V ( T )and W ( T ) related? V,W ( id ) · W ( T ) · W,V ( id ) = ( id ) · ( T id ) = ( id T id ) = V ( T ) ( id ) · ( id ) = ( id ) = 100 ··· 00 010 000 01 V ( T )= P 1 W ( T ) P where P = ( id ) is a change oF basis matrix. On the other hand, iF P is invertible, then P is a change oF basis matrix (see handout). 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Defnition 2 Square matrices A, B are similar if A = P 1 BP for some invertible matrix P . Theorem 3 Suppose that X is fnite-dimensional. IF T L ( X, X ) and U, W are any two bases oF X ,then Mtx W ( T ) and U ( T ) are similar. Conversely, given similar matrices A, B with A = P 1 and any basis U , there is a basis W and T L ( ) such that B = U ( T ) A = W ( T ) P = U,W ( id ) P 1 = W,U ( id ) ProoF: See Handout on Diagonalization and Quadratic Forms. Section 3.6: Eigenvalues and Eigenvectors De la Fuente de±nes eigenvalues and eigenvectors of a matrix. Here, we de±ne eigenvalues and eigenvectors of a linear transfor- mation and show that λ is an eigenvalue of T if and only if λ is an eigenvalue for some matrix representation of T if and only if λ is an eigenvalue for every matrix representation of T . Defnition 4 Let X be a vector space and T L ( ). We say that λ is an eigenvalue of T and v 6 =0i san eigenvector corresponding to λ if T ( v )= λv . Theorem 5 (Theorem 4 in Handout) Let X be a fnite- dimensional vector space, and U any basis. Then λ is an 2
Background image of page 2
eigenvalue of T if and only if λ is an eigenvalue of Mtx U ( T ) . v is an eigenvector of T corresponding to λ if and only if crd U ( v ) is an eigenvector of U ( T ) corrresponding to λ . Proof: By the Commutative Diagram Theorem, T ( v )= λv crd U ( T ( v )) = crd U ( λv ) U ( T )( crd U ( v )) = λ ( crd U ( v )) Computing eigenvalues and eigenvectors: Suppose dim X = n ;le t I be the n × n identity matrix. Given T L ( X, X ), fx a basis U and let A = U ( T ) Find the eigenvalues o± T by computing the eigenvalues o± A : Av = λv ( A λI ) v =0 ( A λI ) is not invertible det( A λI )=0 We have the ±ollowing ±acts: A R n × n , f ( λ )=det( A λI ) is an n th degree polynomial in λ with real coefficients; **it is called the characteristic polynomial A . f has n roots in C , counting multiplicity: f ( λ )=( λ c 1 )( λ c 2 ) ··· ( λ c n ) where c 1 ,...,c n C are the eigenvalues; the c j ’s are not necessarily distinct.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/06/2009 for the course ECON 0204 taught by Professor Staff during the Summer '08 term at University of California, Berkeley.

Page1 / 15

204Lecture102006 - Economics 204 Lecture 10Friday, August...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online