204Lecture102006

# 204Lecture102006 - Economics 204 Lecture 10Friday Revised...

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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 finite-dimensional vector space with basis V . If T L ( X, X ) it is customary to use the same basis in the domain and range: Definition 1 Mtx V ( T ) denotes Mtx V,V ( T ) Question: If W is another basis for X , how are Mtx V ( T ) and Mtx W ( T ) related? Mtx V,W ( id ) · Mtx W ( T ) · Mtx W,V ( id ) = Mtx V,W ( id ) · Mtx W,V ( T id ) = Mtx V,V ( id T id ) = Mtx V ( T ) Mtx V,W ( id ) · Mtx W,V ( id ) = Mtx V,V ( id ) = 1 0 0 · · · 0 0 0 1 0 · · · 0 0 0 0 0 · · · 0 1 Mtx V ( T ) = P 1 Mtx W ( T ) P where P = Mtx W,V ( 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

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Definition 2 Square matrices A, B are similar if A = P 1 BP for some invertible matrix P . Theorem 3 Suppose that X is finite-dimensional. If T L ( X, X ) and U, W are any two bases of X , then Mtx W ( T ) and Mtx U ( T ) are similar. Conversely, given similar matrices A, B with A = P 1 BP and any basis U , there is a basis W and T L ( X, X ) such that B = Mtx U ( T ) A = Mtx W ( T ) P = Mtx U,W ( id ) P 1 = Mtx W,U ( id ) Proof: See Handout on Diagonalization and Quadratic Forms. Section 3.6: Eigenvalues and Eigenvectors De la Fuente defines eigenvalues and eigenvectors of a matrix. Here, we define 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 . Definition 4 Let X be a vector space and T L ( X, X ). We say that λ is an eigenvalue of T and v = 0 is an eigenvector corresponding to λ if T ( v ) = λv . Theorem 5 (Theorem 4 in Handout) Let X be a finite- dimensional vector space, and U any basis. Then λ is an 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 Mtx U ( T ) corrresponding to λ . Proof: By the Commutative Diagram Theorem, T ( v ) = λv crd U ( T ( v )) = crd U ( λv ) Mtx U ( T )( crd U ( v )) = λ ( crd U ( v )) Computing eigenvalues and eigenvectors: Suppose dim X = n ; let I be the n × n identity matrix. Given T L ( X, X ), fix a basis U and let A = Mtx U ( T ) Find the eigenvalues of T by computing the eigenvalues of A : Av = λv ( A λI ) v = 0 ( A λI ) is not invertible det( A λI ) = 0 We have the following facts: If 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 of A .

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