EEL3105_MT2_Solution

# Bweknowthatthedeterminantistheproductofeigenvaluesandm

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

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

Unformatted text preview: 4b 6 0 3a 5b 8 0 We can use any two of these to solve for a and b to get a=‐1, b=‐1. 6. a. Again using the formula that trace equals sum of eigenvalues, we get that the other eigenvalue of A must be 4. b. We know that the determinant is the product of eigenvalues and must therefore be 4. On the other hand, det(A)=6‐b2. Therefore, b2=2 and b= 2. 7. Let’s solve B part first. We are told that A is an nxn matrix whose characteristic polynomial is s n n 1s n 1 n 2 s n 2 1s 0 We start by writing characteristic polynomial of 2A: s s det( sI 2 A) det 2 I A 2n det I A . 2 2 Let us define a new variable z=s...
View Full Document

## This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.

Ask a homework question - tutors are online