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Unformatted text preview: Math 235 Assignment 7 Solutions 1. 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, y ) = 2 x 2 + 6 xy + 2 y 2 . Solution: We have A = bracketleftbigg 2 3 3 2 bracketrightbigg so A I = bracketleftbigg 2 3 3 2 bracketrightbigg . The characteristic equation is 0 = det( A I ) = 2 4 5 = ( 5)( + 1) . The roots are 5 and 1, so these are the eigenvectors of A . For = 5 we get A I = bracketleftbigg 3 3 3 3 bracketrightbigg bracketleftbigg 1 1 bracketrightbigg . Thus, bracketleftbigg 1 1 bracketrightbigg is a corresponding eigenvector. For = 1 we get A I = bracketleftbigg 3 3 3 3 bracketrightbigg bracketleftbigg 1 1 0 0 bracketrightbigg . Thus, bracketleftbigg 1 1 bracketrightbigg is a corresponding eigenvector. Thus, by theorem, braceleftbiggbracketleftbigg 1 1 bracketrightbigg , bracketleftbigg 1 1 bracketrightbiggbracerightbigg is orthogonal. Normalizing the vectors, we get P = 1 2 bracketleftbigg 1 1 1 1 bracketrightbigg , and Q = 5 x 2 1 y 2 1 , where bracketleftbigg x 1 y 1 bracketrightbigg = P T bracketleftbigg x y bracketrightbigg . Since one eigenvalue of A is negative and the other is positive, Q ( x, y ) is indefinite. b) Q ( x, y ) = x 2 + 4 xy + y 2 Solution: We have A = bracketleftbigg 1 2 2 1 bracketrightbigg so A I = bracketleftbigg 1 2 2 1 bracketrightbigg . The characteristic equation is 0 = det( A I ) = 2 2 3 = ( 3)( + 1) . The roots are 3 and 1, so these are the eigenvectors of A . For = 3 we get A I = bracketleftbigg 2 2 2 2 bracketrightbigg bracketleftbigg 1 1 bracketrightbigg . Thus, bracketleftbigg 1 1 bracketrightbigg is a corresponding eigenvector. For = 1 we get A I = bracketleftbigg 2 2 2 2 bracketrightbigg bracketleftbigg 1 1 0 0 bracketrightbigg . Thus, bracketleftbigg 1 1 bracketrightbigg is a corresponding eigenvector....
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 Spring '08
 CELMIN
 Math

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