CS
hw3_solutions

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CS205 Homework #3 Solutions Problem 1 Consider an n × n matrix A . 1. Show that if A has distinct eigenvalues all the corresponding eigenvectors are linearly independent. 2. Show that if A has a full set of eigenvectors (i.e. any eigenvalue λ with multiplicity k has k corresponding linearly independent eigenvectors), it can be written as A = QΛQ - 1 where Λ is a diagonal matrix of A ’s eigenvalues and Q ’s columns are A ’s eigenvectors. Hint: show that AQ = and that Q is invertible. 3. If A is symmetric show that any two eigenvectors corresponding to different eigenvalues are orthogonal. 4. If A is symmetric show that it has a full set of eigenvectors. Hint: If ( λ , q ) is an eigenvalue, eigenvector ( q normalized) pair and λ is of multiplicity k > 1, show that A - λ qq T has an eigenvalue of λ with multiplicity k - 1. To show that consider the Householder matrix H such that Hq = e 1 and note that HAH - 1 = HAH and A are similar. 5. If A is symmetric show that it can be written as A = QΛQ T for an orthogonal matrix Q . (You may use the result of (4) even if you didn’t prove it) Solution 1. Assume that the eigenvectors q 1 , q 2 , . . . , q n are not linearly independent. Therefore, among these eigenvectors there is one (say q 1 ) which can be written as a linear combi- nation of some of the others (say q 2 , . . . , q k ) where k n . Without loss of generality, we can assume that q 2 , . . . , q k are linearly independent (we can keep removing vectors from a linearly dependent set until it becomes independent), therefore the decomposi- tion of q 1 into a linear combination q 1 = k i =2 α k q k is unique. However q 1 = k i =2 α k q k Aq 1 = A k i =2 α k q k λ 1 q 1 = k i =2 λ k α k q k q 1 = k i =2 λ k λ 1 α k q k In the third equation above, we assumed that λ 1 = 0. This is valid, since otherwise the same equation would indicate that there is a linear combination of q 2 , . . . , q k that is equal to zero (which contradicts their linear independence). From the last equation above and the uniqueness of the decomposition of q 1 into a linear combination of q 2 , . . . , q k we have λ k λ 1 α k = α k . There must be an α k 0 = 0 , 2 k 0 k , otherwise we would have q 1 = 0 . Thus λ 1 = λ k 0 which is a contradiction. 1

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2. Since A has a full set of eigenvectors, we can write Aq i = λ i q i , i = 1 , . . . , n . We can collect all these equations into the matrix equation AQ = , where Q has the n eigenvectors as columns, and Λ = diag ( λ 1 , . . . , λ n ). From (1) we have that the columns of Q are linearly independent, therefore it is invertible. Thus AQ = A = QΛQ - 1 . 3. If λ 1 , λ 2 are two distinct eigenvalues and q 1 , q 2 are their corresponding eigenvectors (there could be several eigenvectors per eigenvalue, if they have multiplicity higher than 1), then q T 1 Aq 2 = q T 1 ( Aq 2 ) = λ 2 ( q T 1 q 2 ) q T 1 Aq 2 = q T 1 A T q 2 = ( Aq 1 ) T q 2 = λ 1 ( q T 1 q 2 ) λ 1 ( q T 1 q 2 ) = λ 2 ( q T 1 q 2 ) ( λ 1 - λ 2 )( q T 1 q 2 ) = 0 λ 1 = λ 2 = q T 1 q 2 = 0 4. In the review session we saw that that for a symmetric matrix A , if λ
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• Fall '07
• Fedkiw
• Linear Algebra, linearly independent eigenvectors, ann − λ

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