Q7ts - Quiz VII 10 1. The matrix A = 0 2 10 eigenvectors...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Quiz VII 1. The matrix A = 1 0 1 0 2 0 1 0 1 is symmetric. Find a basis for R 3 consisting of eigenvectors for A . Solution: Since A is symmetric, there will be an orthonormal basis consisting of eigenvectors of A , but you do not need to use that to solve the problem. First find the eigenvalues det 1 - λ 0 1 0 2 - λ 0 1 0 1 - λ = (1 - λ ) 2 (2 - λ ) - (2 - λ ) = (2 - λ )( - 2 λ + λ 2 ) . So the eigenvalues are 2 and 0. Solving for the eigenspaces, ⃗v 1 = 1 0 - 1 is an eigenvector for the eigenvalue 0. For the eigenvalue 2, A - 2 I = - 1 0 1 0 0 0 1 0 - 1 , and the reduced row echelon form of that is just 1 0 - 1 0 0 0 0 0 0 So a basis for the eigenspace E 2 is { ⃗v 2 = 0 1 0 ,⃗v 3 = 1 0 1 } . The basis B = { ⃗v 1 ,⃗v 2 ,⃗v 3 } is a basis of eigenvectors. Note that, if you divide ⃗v 1 and ⃗v 3 by their lengths (both lengths are 2), this becomes an orthonormal basis. 2. Find a basis for R 2 such that the matrix of T ( ⃗x ) = ( 3 2 - 1 1 ) ⃗x with respect to this basis has the form
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

Q7ts - Quiz VII 10 1. The matrix A = 0 2 10 eigenvectors...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online