Question

Let V be a finite-dimensional inner product space. For an operator TEL(V), define its norm by ||T|:= max{||Tull VEV. ||0|| = 1}. (1) To explain this, note that {l|Tu ve V, || 0 || = 1} is a non-empty subset of [0,00). The expression max{||TV|| | V EV, ||0|| = 1} means the maximum, or largest, value in this set. In words, the norm of an operator describes the maximal amount that it 'stretches' (or shrinks) vectors. (a) (1 point) What is the norm of the identity operator? (b) (2 points) Prove that the expression in line (1) above defines a norm on L(V) in the sense of Homework 7, question 2. (c) (1 point) Prove that for any TEL(V), ||1|| > max{[X] is an eigenvalue of T}.

Answer 1

$\text{(a) The identity operator takes a vector to itself. }Iv=v. \text{ hence } ||I||=$ $\text{(a) The identity operator takes a vector to itself. }Iv=v. \text{ hence } ||I||=max\{||Iv||: ||v||=1, v\in V\}=max\{||v||:||v||=1, v\in V\}=1\\ \text{(c) Let }\lambda \text{ be an eigenvalue of T, with corr eigenvector v. }Tv=\lambda v. \Rightarrow ||Tv||=||\lambda v||=|\lambda|||v||\\ ||Tv||\leq ||T||||v||, \text{ if }||v||=1, ||T||\geq |\lambda|, \Rightarrow ||T||\geq max\{|\lambda|: \lambda \text{ an eigenvalue of }T\}\\ \text{(d) T is normal, hence by Spectral Theorem, there exists an Orthonormal basis of V consisting of the eigenvectors. Let }v\in V, v_1, v_2, .... \text{ be eigenvectors with eigenvalues }\lambda_1, \lambda_2,.....\\ v= c_1v_1+....., \Rightarrow Tv=\lambda_1c_1v_1+....., ||Tv||^2=\langle Tv, Tv \rangle=\lambda_1^2c_1^2+.... \text{ (As the eigenvectors are orthonomral with norm 1.)}\\ ||v||^2=c_1^2+.. \Rightarrow \frac{||Tv||^2}{||v||^2}=\frac{\lambda_1^2c_1^2+....}{c_1^2+..}\leq max\{\lambda_1^2,.. \}$ Take ||v||=1, by square root, ||T|| <= max{|lambda| : lambda is an eigenvalue of T}. Together with previous part, equality is achieved.

(b) Info on HW problem not provided