Unformatted text preview: Math 330 Quiz 5.2
Determine the eigenvalues and eigenspaces of the matrix A: 3 0 0 A= 0 1 2 0 2 1 Is there a basis for R2 consisting of eigenvectors for A? SOLUTION: A  I = 3 0 0 0 1 2 0 2 1 = (3  ) 1 2 2 1 = (3  ) (1  )2  4 = (3  ) 2  2 + 1  4 = (3  )(2  2  3) = (3  )(  3)( + 1) So the eigenvalues are = 3 and = 1. To find the eigenspaces we proceed with each eigenvalue separately: = 1
We examine the null space of A + I: 4 0 0 (R1,R2)( R1 , R2 ) 1 0 0 4 0 0 R3R3R2 4 2 0 2 2 0 1 1 = = A+I = 0 2 2 0 0 0 0 0 0 0 2 2 x1 So with an associated eigenvector of x = x2 we have that x3 x1 0 x1 = 0 x2 = x3 1 x2 = x3 x3 1 0 So the 1eigenspace is the span of 1 . 1 =3
Here we look at the null space of A  3I: 0 0 0 0 0 0 R2 R2 0 0 0 R3R3+R2 2 0 2 2 = 0 1 1 A  3I = 0 2 2 = 0 2 2 0 0 0 0 0 0 x1 So our eigenvector x = x2 must satisfy just the single equation x2 = x3 . Thus both x1 and x3 are free x3 variables and hence the 3eigenspace is the span of 0 1 0 , 1 0 1 So we do indeed have a basis for R3 consisting of eigenvectors for A. 1 ...
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 Spring '08
 Johnson,J
 Linear Algebra, Algebra, Eigenvectors, Vectors, Eigenvalue, eigenvector and eigenspace, Singular value decomposition, R2 R2

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