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

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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...
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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.

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

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