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Unformatted text preview: Math 115A Homework 4 Due November 21st, 2008 1. Given the field F , the Fvector space V and the linear transformation T : V → V , compute all the eigenvalues and eigenvectors of T . (2 pts each) a) F = R , V = C ∞ ( R ) the space of infinitely differentiable functions, and T : V → V the second derivative, that is, T ( f )( x ) = f 00 ( x ). All real numbers occur as eigenvalues. If λ > 0, then the eigenvectors with eigenvalue λ are just functions of the form c 1 e √ λx + c 2 e √ λx for scalars c 1 and c 2 . If λ = 0, then the eigenvectors are linear polynomials. Finally, if λ < 0, then the eigenvectors are functions of the form c 1 sin( √ λx ) + c 2 cos( √ λx ), for scalars c 1 and c 2 . (Note the eigenspaces are all of dimension two, but they look very differently depending on whether λ is positive, 0, or negative.) b) F = C , V = P 3 the space of complex polynomials of degree at most 3, and T : V → V given by T ( f )( z ) = f ( z ) + f ( z ). The only eigenvalue is 1. The eigenspace E 1 consists of the constant polynomials. 2. Let F be a field, V an Fvector space and T : V → V a linear transformation such that there exists a natural number k with T k = 0. Determine all the eigenvalues of T . ( Remark: Such a linear transformation is called nilpotent.) (4 pts) If λ ∈ F is an eigenvalue of T , then λ k is an eigenvalue of T k . Since T k = 0, we have that λ k = 0; because F is a field, this implies λ = 0. Therefore the only possible eigenvalue of T is 0. I claim that 0 actually occurs as an eigenvalue, that is, that N ( T ) 6 = { } . Indeed, let n be the minimal positive integer such that T n = 0. Then there exists v ∈ V such that T n 1 ( v ) 6 = 0 (otherwise n wasn’t minimal). On the other hand, T ( T n 1 ( v )) = T n ( v ) = 0, so that T n 1 ( v ) ∈ N ( T ) is an eigenvector with eigenvalue 0.) is an eigenvector with eigenvalue 0....
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This note was uploaded on 02/21/2010 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.
 Spring '10
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