This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 17B
Kouba Using Eigenvalues and Eigenvectors to Solve Matrix Algebra Problems DERIVE A FORMULA FOR POWERS OF A MATRIX APPLIED TO ITS EIGENVEC
TOR : Let A be a matrix and assume that A is an eigenvalue with eigenvector V . Then
AV 2 AV —>
A2V = A(AV) = A(AV) = ,\(AV) 2 /\()\V) = AZV _>
A3V = A(A2V) = A(A2V) :2 A2(AV) = A2()\V) = A3V ———>
A4V : A(A3V) = A()\3V) = A3(AV) : A3()\V) : A4V ——+ (P) AkV—zxkv for k:1,2,3,4, USE THE ABOVE FORMULA TO SOLVE THE FOLLOWING PROBLEM : EXAMPLE : Consider matrix A = (—11/2 3721) . It can be shown that it’s eigenvalues
are /\1 = 2 and A2 = —1 with eigenvectcors V1 : (i) and V2 2 (35 ), resp. Use the eigenvalues and eigenvectors for matrix A to determine A30 (123) . Begin by writing vector (123) as a linear combination of the eigenvectors <1) and <25): 2
<23>=<m><a>+<ca><25>  13: C1 :— 562
2:201+202 —> ~26 : —~201 + 1002
2=2cl+202 —> —24:12C2 —> 022—2 and c1=3. Thus,wehave (123)2(3)(;)+(—2)(‘25) _., 1 ,221,225,482
,442,450,940 ' ...
View
Full Document
 Winter '07
 Kouba
 Algebra

Click to edit the document details