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Unformatted text preview: AMS210.01. Practice Final. Solutions. Andrei Antonenko 1. Find the square root of the following matrix: A = 4 5 3 6 ! Solution: To find a square root of the matrix, we first diagonalize it. So, first lets find its eigenvalues. The characteristic polynomial is p A ( ) = 2 10 + 9 . So, the roots of the characteristic polynomial are 1 = 1, and 2 = 9. Thus, the diagonal form of the matrix is D = 1 0 0 9 ! . Now, lets find corresponding eigenvectors: 1 = 1 . Subtracting 1 = 1 from the diagonal of the given matrix we get: A 1 I = 3 5 3 5 ! . So, the equation to determine the eigenvector is 3 x + 5 y = 0 , thus an eigenvector is (5 , 3). 2 = 9 . Subtracting 1 = 1 from the diagonal of the given matrix we get: A 1 I =  5 5 3 3 ! . So, the equation to determine the eigenvector is 5 x + 5 y = 0 , thus an eigenvector is (1 , 1). 1 Now, writing eigenvectors as columns of the matrix C , we have C = 5 1 3 1 ! , C 1 = 1 8 1 1 3 5 ! . For such defined matrices D,C , and C 1 we have A = CDC 1 . Now, the square root of the diagonal form of A is D = 1 3 ! Lets take one of possible values of D : D = 1 0 0 3 ! . Thus, square root of A would be: A = C DC 1 = 1 8 5 1 3 1 ! 1 0 0 3 ! 1 1 3 5 ! = 1 8 5 3 3 3 ! 1 1 3 5 ! = 1 8 14 10 6 18 ! = 1 4 7 5 3 9 ! . Taking other possible values of D we can get other square roots of A . 2. Find the formula for the mth power of the matrix A = 2 1 1 2 ! Solution: Lets use approximation polynomial method. But first we have to find eigenvalues. The characteristic equation is p A ( ) = 2 4 + 3 = 0 , thus the eigenvalues are 1 = 1, and 2 = 3. Now the approximation equation is at + b = t m . Substituting different values of into this equation we get a system with 2 unknowns a and b : ( a + b = 1 3 a + b = 3 m 2 Lets solve it. Subtracting the first equation from the second one we get 2 a = 3 m 1 , so a = 3 m 1 2 . Now, b = 1 a = 1 3 m 1 2 = 3 3 m 2 . Now, the matrix form of the approximation equation is aA + bI = A m . Substituting a,b and A , we get: A m = 3 m 1 2 2 1 1 2 ! + 3 3 m 2 1 0 0 1 ! = 1 2 2(3 m 1) + (3 3 m ) 3 m 1 3 m 1 2(3 m 1) + (3 3 m ) ! = 1 2 3 m + 1 3 m 1 3 m 1 3 m + 1 ! . 3. Find the distance between the vector u = (10 , 5 , 12) and the subspace, spanned by the following vectors: v 1 = (4 , 1 , 2); v 2 = (5 , 8 , 7); v 3 = (6 , 3 , 0) . Solution: First we should apply the GramSchmidt orthogonalization process to vectors v 1 ,v 2 , and v 3 : w 1 = v 1 = (4 , 1 , 2); w 2 = v 2 h v 3 ,w 1 i h w 1 ,w 1 i w 1 = (5 , 8 , 7) 20 + 8 + 14 16 + 1 + 4 (4 , 1 , 2) = (5 , 8 , 7) 2(4 , 1 , 2) = ( 3 , 6 , 3); w 3 = v 3 h v 3 ,w 1 i h w 1 ,w 1 i w 1 h v 3 ,w 2 i h w 2 ,w 2 i w 2 = (6 , 3 , 0) 24 3 16 + 1 + 4 (4 , 1 , 2) 18 18 9 + 36 + 9 ( 3 , 6 , 3) = (6 , 3 , 0) (4 , 1 , 2) + 2 3 ( 3 , 6 , 3) = (0 , , 0) ....
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.
 Spring '03
 Andant

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