A8_soln - Math 235 Assignment 8 Solutions 1 Let A = 1 − 2...

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Unformatted text preview: Math 235 Assignment 8 Solutions 1. Let A = 1 − 2 3 1 2 . Find the maximum and minimum value of bardbl Avectorx bardbl for vectorx ∈ R 2 subject to the constraint bardbl vectorx bardbl = 1. Solution: We have A T A = bracketleftbigg 11 0 8 bracketrightbigg . Hence, the eigenvalues of A T A are λ 1 = 11 and λ 2 = 8. Thus the maximum of bardbl Avectorx bardbl subject to bardbl vectorx bardbl = 1 is √ 11 and the minimum is √ 8. 2. For each quadratic form Q ( vectorx ), determine the corresponding symmetric matrix A . By diagonalizing A , express Q ( vectorx ) in diagonal form and give an orthogonal matrix that diago- nalizes A . Classify each quadratic form. a) Q ( x 1 , x 2 ) = 3 x 2 1 − 4 x 1 x 2 + 6 x 2 2 . Solution: We have A = bracketleftbigg 3 − 2 − 2 6 bracketrightbigg so A − λI = bracketleftbigg 3 − λ − 2 − 2 6 − λ bracketrightbigg . The characteristic polynomial is 0 = det( A − λI ) = λ 2 − 9 λ + 14 = ( λ − 2)( λ − 7) Thus the eigenvalues of A are λ 1 = 2 and λ 2 = 7. For λ 1 = 2 we get A − 2 I = bracketleftbigg 1 − 2 − 2 4 bracketrightbigg ∼ bracketleftbigg 1 − 2 bracketrightbigg Hence, a basis for E λ 1 is braceleftbiggbracketleftbigg 2 1 bracketrightbiggbracerightbigg . For λ = 7 we get A − 7 I = bracketleftbigg − 4 − 2 − 2 − 1 bracketrightbigg ∼ bracketleftbigg 1 1 / 2 bracketrightbigg Hence, a basis for E λ 2 is braceleftbiggbracketleftbigg − 1 2 bracketrightbiggbracerightbigg ....
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This note was uploaded on 01/21/2012 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.

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A8_soln - Math 235 Assignment 8 Solutions 1 Let A = 1 − 2...

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