mat67-Homework6_Solutions

# mat67-Homework6_Solutions - MAT067 University of California...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAT067 University of California, Davis Winter 2007 Solutions to Homework Set 6 As usual, we are using F to denote either R or C , and F [ z ] denotes the set of polynomials with coefficients over F . 1. Let V be a finite-dimensional vector space over F , and let S, T ∈ L ( V ) be linear operators on V with S invertible. Given any polynomial p ( z ) ∈ F [ z ], prove that p ( S ◦ T ◦ S- 1 ) = S ◦ p ( T ) ◦ S- 1 . Solution : Proposition. Let V be a finite-dimensional vector space over F with S, T ∈ L ( V ) linear operators on V such that S invertible. Given p ( z ) ∈ F [ z ] a polynomial, p ( S ◦ T ◦ S- 1 ) = S ◦ p ( T ) ◦ S- 1 . Proof. Let p ( z ) ∈ F [ z ] be a polynomial with coefficients in F , and suppose the p ( z ) = ∑ n k =0 a k z k where a k ∈ F and n = deg( p ). Then, note that for each monomial a k z k in p , the corresponding monomial in p ( S ◦ T ◦ S- 1 ) is a k ( S ◦ T ◦ S- 1 ) k = a k ( S ◦ T ◦ S- 1 ) · ··· · ( S ◦ T ◦ S- 1 ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright k times = ··· = a k S ◦ T k ◦ S- 1 , where we have repeatedly used the fact that S- 1 ◦ S is the identity operator on V . Thus, it follows that p ( S ◦ T ◦ S- 1 ) can equivalently be written as n summationdisplay k =0 a k ( S ◦ T ◦ S- 1 ) k = n summationdisplay k =0 a k S ◦ T k ◦ S- 1 = S ◦ parenleftBigg n summationdisplay k =0 a k T k parenrightBigg ◦ S- 1 = S ◦ p ( T ) ◦ S- 1 . 2 2. Let V be a finite-dimensional vector space over C , T ∈ L ( V ) be a linear operator on V , and p ( z ) ∈ C [ z ] be a polynomial. Prove that λ ∈ C is an eigenvalue of the linear operator p ( T ) ∈ L ( V ) if and only if T has an eigenvalue μ ∈ C such that p ( μ ) = λ . Solution : Proposition. Let V be a finite-dimensional vector space over C and T ∈ L ( V ) be a linear operator on V . Then, given any polynomial p ( z ) ∈ C [ z ] , λ ∈ C is an eigenvalue of the linear operator p ( T ) ∈ L ( V ) if and only if T has an eigenvalue μ ∈ C such that p ( μ ) = λ ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

mat67-Homework6_Solutions - MAT067 University of California...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online