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Unformatted text preview: Chapter 4. Eigenvalues and Eigenvectors. In this chapter V is an ndimensional vector space over F = R or C . Recall L ( V ) is the vector space of all linear maps from V to V . Let f L ( V ). An eigenvalue for f is an element a F such that there exists a nonzero vector v V with f ( v ) = av . For a F , the eigenspace for a (on V with respect to f ) is E ( a ) = { v V : f ( v ) = av } . Thus a is an eigenvalue of f iff E ( a ) 6 = 0. The nonzero members of E ( a ) are the eigenvectors for the eigenvalue a . The eigenspace E ( a ) is a subspace of V ; indeed: Lemma 4A. Let f L ( V ) and a F . Then E ( a ) = N ( a id V f ) . Proof. Recall id V is the identity map on V and id V L ( V ). As L ( V ) is a vector space and f and id V are in L ( V ), so is g = a id V f . Further by definition of addition and scalar multiplication in the vector space L ( V ), g ( v ) = a id V ( v ) f ( v ) = av f ( v ) , so g ( v ) = 0 iff f ( v ) = av . That is N ( g ) = E ( a ) as claimed. Examples (1) Notice f = 0 id V f and N ( f ) = N ( f ), so by Lemma 4A, E (0) = N ( f ). (2) Take f = id V to be the identity map on V . Then for all 0 6 = v V , v = f ( v ) = 1 v , so v is an eigenvector for f with eigenvalue 1. Thus 1 is an eigenvalue for the identity map and V = E (1) is the eigenspace for this eigenvalue. Theorem 4.2. Let a 1 ,... ,a m be distinct eigenvalues for f L ( V ) and v i an eigenvector for a i . Then { v 1 ,... ,v m } is linearly independent of order m . Proof. The proof is by induction on m . If m = 1 the result holds as eigenvectors are nonzero. Thus we may take m &gt; 1, and proceeding by induction on m , assume that each proper subset of Y = { v 1 ,... ,v m } is independent. Suppose 0 = i b i v i is a dependence relation on Y . As each proper subset of Y is independent, each b j is nonzero. Now 0 = f ( X i b i v i ) = X i b i f ( v i ) = X i b i a i v i . Also 0 = a m ( i b i v i ) = i a m b i v i , so 0 = X i ( a i a m ) b i v i = X i&lt;m c i v i , where c i = ( a i a m ) b i , since c m = 0. However for j &lt; m , a m 6 = a j , so c j 6 = 0. Thus we have a dependence relation on { v 1 ,... ,v m 1 } , contrary to an earlier observation. 1 2 Polynomial functions of matrices . Let x be a symbol and define a polynomial in x over F to be a formal sum f ( x ) = m X i =0 a i x i , for some nonnegative integer m and elements a i F . Formally a polynomial is just a sequence of elements of F such that all but a finite number of terms in the sequence are 0; that is i a i x i denotes the sequence a ,a 1 ,... . The polynomial notation is one way to write such sequences. Call a i the i th coefficient of the polynomial f . The zero polynomial is the polynomial all of whose coefficients are 0. The degree of the zero polynomial is 0, and for all other polynomials f ( x ) = i a i x i we define the degree of f to be deg( f ) = max { i : a i 6 = 0 } ....
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This note was uploaded on 03/03/2011 for the course MATH 1B taught by Professor Aschbacher during the Winter '07 term at Caltech.
 Winter '07
 Aschbacher
 Linear Algebra, Algebra, Eigenvectors, Vectors, Vector Space

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