{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Macroeconomics Exam Review 130

# Macroeconomics Exam Review 130 - Solutions for Foundations...

This preview shows page 1. Sign up to view the full content.

3. Since 𝑓 is symmetric 𝑔 ( y , x ) = ( 𝜆 y 𝑓 ( y ) ) 𝑇 x = 𝜆 y 𝑇 x 𝑓 ( y ) 𝑇 x = 𝜆 x 𝑇 y 𝑓 ( x ) 𝑇 𝑦 = ( 𝜆 x 𝑓 ( x ) ) 𝑇 y = 𝑔 ( x , y ) 4. 𝑔 satisfies the conditions of Exercise 3.59 and therefore ( 𝑔 ( x , y )) 2 𝑔 ( x , x ) 𝑔 ( y , y ) for every x , y 𝑋 (3.41) By definition 𝑔 ( x 0 , x 0 ) = 0 and (3.41) implies that 𝑔 ( x 0 , y ) = 0 for every y 𝑋 That is 𝑔 ( x 0 , y ) = ( 𝜆 x 0 𝑓 ( x 0 ) ) 𝑇 y = 0 for every 𝑦 𝑋 and therefore 𝜆 x 0 𝑓 ( x 0 ) = 0 or 𝑓 ( x 0 ) = 𝜆 x 0 In other words, x 0 is an eigenvector. By construction, x 0 = 1. 3.88 1. Suppose x 2 , x 3 𝑆 . Then ( 𝛼 x 2 + 𝛽 x 3 ) 𝑇 x 1 = 𝛼 x 𝑇 2 x 1 + 𝛽 x 𝑇 3 x 1 = 0 so that 𝛼 x 2 + 𝛽 x 3 𝑆 . 𝑆 is a subspace. Let { x 1 , x 2 , . . . , x 𝑛 } be a basis for 𝑋 (Exercise 1.142). For x 𝑋 , there exists (Exercise 1.137) unique 𝛼 1 , 𝛼 2 , . . . , 𝛼 𝑛 such that x = 𝛼 1 x 1 + 𝛼 2 x 2 + ⋅ ⋅ ⋅ + 𝛼 𝑛 x 𝑛 If x 𝑆 x 𝑇 x 1 = 𝛼 1 x 𝑇 1 x 1 = 0
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online