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PrFinalSol

# PrFinalSol - AMS210.01 Practice Final Solutions Andrei...

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AMS210.01. Practice Final. Solutions. Andrei Antonenko 1. Find the square root of the following matrix: A = ˆ 4 5 3 6 ! Solution: To find a square root of the matrix, we first diagonalize it. So, first let’s find its eigenvalues. The characteristic polynomial is p A ( λ ) = λ 2 - 10 λ + 9 . So, the roots of the characteristic polynomial are λ 1 = 1, and λ 2 = 9. Thus, the diagonal form of the matrix is D = ˆ 1 0 0 9 ! . Now, let’s find corresponding eigenvectors: λ 1 = 1 . Subtracting λ 1 = 1 from the diagonal of the given matrix we get: A - λ 1 I = ˆ 3 5 3 5 ! . So, the equation to determine the eigenvector is 3 x + 5 y = 0 , thus an eigenvector is (5 , - 3). λ 2 = 9 . Subtracting λ 1 = 1 from the diagonal of the given matrix we get: A - λ 1 I = ˆ - 5 5 3 - 3 ! . So, the equation to determine the eigenvector is - 5 x + 5 y = 0 , thus an eigenvector is (1 , 1). 1

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Now, writing eigenvectors as columns of the matrix C , we have C = ˆ 5 1 - 3 1 ! , C - 1 = 1 8 ˆ 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 = ˆ ± 1 0 0 ± 3 ! Let’s take one of possible values of D : D = ˆ 1 0 0 3 ! . Thus, square root of A would be: A = C DC - 1 = 1 8 ˆ 5 1 - 3 1 ! ˆ 1 0 0 3 ! ˆ 1 - 1 3 5 ! = 1 8 ˆ 5 3 - 3 3 ! ˆ 1 - 1 3 5 ! = 1 8 ˆ 14 10 6 18 ! = 1 4 ˆ 7 5 3 9 ! . Taking other possible values of D we can get other square roots of A . 2. Find the formula for the m -th power of the matrix A = ˆ 2 1 1 2 ! Solution: Let’s use approximation polynomial method. But first we have to find eigenvalues. The characteristic equation is p A ( λ ) = λ 2 - 4 λ + 3 = 0 , thus the eigenvalues are λ 1 = 1, and λ 2 = 3. Now the approximation equation is at + b = t m . Substituting different values of λ into this equation we get a system with 2 unknowns a and b : ( a + b = 1 3 a + b = 3 m 2
Let’s 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 ˆ 2 1 1 2 ! + 3 - 3 m 2 ˆ 1 0 0 1 ! = 1 2 ˆ 2(3 m - 1) + (3 - 3 m ) 3 m - 1 3 m - 1 2(3 m - 1) + (3 - 3 m ) ! = 1 2 ˆ 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 Gram-Schmidt 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 , 0) . 3

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So, we have: w 1 = (4 , 1 , 2); w 2 = ( - 3 , 6 , 3); w 3 = (0 , 0 , 0) . Now we can find the projection of u = (10 , 5 , - 12) along the subspace spanned by w 1 , w 2 , and w 3 . But w 3 = 0, so we simply omit it: proj( u ) = h u, w 1 i h w 1 , w 1 i w 1 + h u, w 2 i h w 2 , w 2 i w 2 = 40 + 5 - 24 16 + 1 + 4 (4 , 1 , 2) + - 30 + 30 - 36 9 + 36 + 9 ( - 3 , 6 , 3) = (4 , 1 , 2) - 2 3 ( - 3 , 6 , 3) = (6 , - 3 , 0) .
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PrFinalSol - AMS210.01 Practice Final Solutions Andrei...

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