# prac05s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

This preview shows pages 1–3. Sign up to view the full content.

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

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

Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Practice session 5 — Solutions 1. Consider the following two matrices: A = parenleftBig 0 0- 2 1 2 1 1 0 3 parenrightBig and B = parenleftBig 3 1 2 4 parenrightBig . For each of these matrices: a ) Find all the eigenvalues for the matrix and, for each eigenvalue, find a basis for the corre- sponding eigenspace. b ) If possible, find a basis of R 3 for A , and of R 2 for B , consisting of eigenvectors. c ) Write down the square matrix P whose columns are the basis vectors you found in part (b). (Such a matrix is invertible. Why?) Write down the diagonal matrix D whose diagonal entries are the eigenvalues of the matrix, in the same order as the corresponding columns of P . d ) Check by doing the multiplications that AP = PD . (This confirms that A is diagonalis- able; that is, P- 1 AP = D .) Solution (I) A = parenleftBig 0 0- 2 1 2 1 1 0 3 parenrightBig a ) To find the eigenvalues of A we solve the equation det( A- λI ) = 0 : vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle- λ- 2 1 2- λ 1 1 3- λ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle =- λ (2- λ )(3- λ )- 2(- (2- λ )) = (2- λ )( λ 2- 3 λ + 2) = (2- λ )( λ- 2)( λ- 1) Hence, the eigenvalues of A are 2 and 1. The eigenspace for λ = 1 : We solve the equation ( A- I ) v = for v : parenleftBigg- 1- 2 1 1 1 1 2 parenrightBigg Row reduce-----→ parenleftBigg 1 2 1- 1 parenrightBigg Therefore v = parenleftBig- 2 t t t parenrightBig , for t ∈ R . Hence, a basis for the 1-eigenspace of A is braceleftBigparenleftBig- 2 1 1 parenrightBigbracerightBig . The eigenspace for λ = 2 : We solve the equation ( A- 2 I ) v = for v : parenleftBigg- 2- 2 1 1 1 1 parenrightBigg Row reduce-----→ parenleftBigg 1 1 parenrightBigg . Therefore v = parenleftBig- t s t parenrightBig , for s, t ∈ R . As parenleftBig- t s t parenrightBig = t parenleftBig- 1 1 parenrightBig + s parenleftBig 1 parenrightBig , a basis for the 2 –eigenspace of A is braceleftBigparenleftBig- 1 1 parenrightBig , parenleftBig 1 parenrightBigbracerightBig . Math 2061: Practice session 5 — Solutions A.M. 5/1/2012 Linear Mathematics Practice session 5 — Solutions Page 2 b ) A basis of eigenvectors for R 3 is braceleftBigparenleftBig- 2 1 1 parenrightBig , parenleftBig- 1 1 parenrightBig , parenleftBig 1 parenrightBigbracerightBig . c ) Let P = parenleftBig- 2- 1 0 1 0 1 1 1 0 parenrightBig . Then P is invertible since its columns are linearly independent. d ) By definition the columns of P are eigenvectors for A ; hence, we have that AP = A parenleftBig- 2- 1 0 1 0 1 1 1 0 parenrightBig = parenleftBigg A parenleftBig- 2 1 1 parenrightBig , A parenleftBig- 1 1 parenrightBig , A parenleftBig 1 parenrightBig parenrightBigg = parenleftBigg 1 parenleftBig- 2 1 1 parenrightBig , 2 parenleftBig- 1 1 parenrightBig , 2 parenleftBig 1 parenrightBig parenrightBigg = parenleftBig- 2- 1 0 1 0 1 1 1 0...
View Full Document

## This note was uploaded on 02/06/2012 for the course MATH 2061 taught by Professor Notsure during the Three '09 term at University of Sydney.

### Page1 / 6

prac05s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online