tut05s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Tutorial 5 — Solutions 1. Consider the following two matrices A = parenleftBig 2 0- 2 0 3 0 0 0 3 parenrightBig and B = parenleftBig 1 0 0 0 1 1 0 1 1 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 ) Find a basis of R 3 consisting of eigenvectors of the matrix. 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 ) Confirm that the matrix is diagonalisable by performing appropriate matrix multiplica- tions. Solution (I) A = parenleftBig 2 0- 2 0 3 0 0 0 3 parenrightBig a ) To find the eigenvectors of A we solve the equation | A − λI | = 0 : vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 2 − λ − 2 3 − λ 3 − λ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = (2 − λ )(3 − λ ) 2 . Hence, the eigenvalues are λ = 2 and λ = 3 . Note that the solution λ = 2 occurs with algebraic multiplicity 2. 2-eigenspace: A non-zero vector v ∈ R 3 is a 2 –eigenvector for A if and only if ( A − 2 I ) v = . The corresponding augmented matrix equation is: parenleftBigg − 2 1 1 parenrightBigg Row reduce −−−−−→ parenleftBigg 1 1 parenrightBigg . Hence, v = parenleftBig t parenrightBig , for t ∈ R , t negationslash = 0 , and a basis for the 2-eigenspace is braceleftBigparenleftBig 1 parenrightBigbracerightBig . 3-eigenspace : This time we solve ( A − 3 I ) v = , which corresponds to the following augmented matrix equation: parenleftBigg − 1 − 2 parenrightBigg Row reduce −−−−−→ parenleftBigg 1 2 parenrightBigg . Hence, v = parenleftBig- 2 t s t parenrightBig , for s, t ∈ R (with s and t not both equal to ) and braceleftBigparenleftBig- 2 1 parenrightBig , parenleftBig 1 parenrightBigbracerightBig is a basis for the 3 –eigenspace of A . Note that the dimension of this eigenspace is 2. Math 2061: Tutorial 5 — Solutions A.M. 5/1/2012 Linear Mathematics Tutorial 5 — Solutions Page 2 b ) By the calculations in part(a), an eigenvector basis of R 3 is braceleftBigparenleftBig 1 parenrightBig , parenleftBig- 2 1 parenrightBig , parenleftBig 1 parenrightBigbracerightBig Note that it is obvious that this set is linearly independent. c ) Set P = parenleftBig 1- 2 0 0 0 1 0 1 0 parenrightBig and D = parenleftBig 2 0 0 0 3 0 0 0 3 parenrightBig . d ) By direct calculation AP = parenleftBig 2 0- 2 0 3 0 0 0 3 parenrightBigparenleftBig 1- 2 0 0 0 1 0 1 0 parenrightBig = parenleftBig 2- 6 0 0 0 3 0 3 0 parenrightBig and PD = parenleftBig 1- 2 0 0 0 1 0 1 0 parenrightBigparenleftBig 2 0 0 0 3 0 0 0 3 parenrightBig = parenleftBig 2- 6 0 0 0 3 0 3 0 parenrightBig...
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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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tut05s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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