svd_notes

# svd_notes - CHAPTER 8 Symmetric Matrices and Quadratic...

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CHAPTER 8 Symmetric Matrices and Quadratic Forms Section 8.4 Singular Value Decomposition A common type of problem in economics or the sciences is to find the maximum and/or minimum of a function subject to a constraint. An example of this type of problem is to find the maximum/minimum of a quadratic form Q ( ~x ) subject to the constraint k ~x k = 1 . We now look at how to solve this type of problem and then generalize the method to non-square matrices. Find the maximum and minimum of Q ( x 1 , x 2 ) = 2 x 2 1 + 3 x 2 2 subject to the constraint EXAMPLE 1 k ( x 1 , x 2 ) k = 1 . Solution: Let’s see if we can determine the maximum and minimum value by first trying some points ( x 1 , x 2 ) that satisfy the constraint. Q (1 , 0) = 2 Q 1 2 , 1 2 = 5 2 Q 1 2 , 3 2 ! = 11 4 Q (0 , 1) = 3 Although we have only tried a few points, we may guess that 3 is the maximum value. Indeed, we have that Q ( x 1 , x 2 ) = 2 x 2 1 + 3 x 2 2 3 x 2 1 + 3 x 2 2 = 3( x 2 1 + x 2 2 ) = 3 , since k ( x 1 , x 2 ) k 2 = 1 implies x 2 1 + x 2 2 = 1 . Hence, 3 is an upper bound for Q and Q (0 , 1) = 3 so 3 is the maximum value of Q . Similarly, we have Q ( x 1 , x 2 ) = 2 x 2 1 + 3 x 2 2 2 x 2 1 + 2 x 2 2 = 2( x 2 1 + x 2 2 ) = 2 , and Q (1 , 0) = 2 , so the minimum value is 2 . This example was extremely easy since we had no cross-terms in the quadratic form. We now look at what happens if we have cross-terms.

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2 Chapter 8 Symmetric Matrices and Quadratic Forms Find the maximum and minimum of Q ( x 1 , x 2 ) = 7 x 2 1 + 8 x 1 x 2 + x 2 2 subject to x 2 1 + x 2 2 = 1 . EXAMPLE 2 Solution: We first try some values of ( x 1 , x 2 ) that satisfy the constraint. We have Q (1 , 0) = 7 Q 1 2 , 1 2 = 8 Q (0 , 1) = 1 Q - 1 2 , 1 2 = 0 Q ( - 1 , 0) = 7 From this we might think that the maximum is 8 and the minimum is 0, although we realize that we have tested very few points. Indeed, if we try more points, we quickly find that these are not the maximum and minimum. So, instead of testing points, we need to try to think of where the maximum and minimum should occur. In the previous example, they occurred at (1 , 0) and (0 , 1) which are the principal axis. Thus, it makes sense to consider the principal axis of this quadratic form as well. We find that Q has eigenvalues λ = 9 and λ = - 1 with corresponding unit eigenvalues ~v 1 = 2 / 5 1 / 5 and ~v 2 = 1 / 5 - 2 / 5 . Taking these for ( x 1 , x 2 ) we get Q 2 5 , 1 5 = 9 , Q 1 5 , - 2 5 = - 1 Moreover, taking ~ y = P T ~x = 1 5 2 1 1 - 2 x 1 x 2 = 1 5 2 x 1 + x 2 x 1 - 2 x 2 we get Q = ~ y T D~ y = 9 2 5 x 1 + 1 5 x 2 2 - 1 5 x 1 - 2 5 x 2 2 9 2 5 x 1 + 1 5 x 2 2 + 9 1 5 x 1 - 2 5 x 2 2 = 9 " 2 5 x 1 + 1 5 x 2 2 + 1 5 x 1 - 2 5 x 2 2 # = 9 , since 2 5 x 1 + 1 5 x 2 2 + 1 5 x 1 - 2 5 x 2 2 = x 2 1 + x 2 2 = 1 . Similarly, we can show that Q ≥ - 1 . Thus, The maximum of is 9 and the minimum is - 1 .
Section 8.4 Singular Value Decomposition 3 Let Q ( ~x ) be a quadratic form on R n with corresponding symmetric matrix A . The maxi- Theorem 1 mum value and minimum value of Q subject to the constraint k ~x k = 1 is the greatest and least eigenvalue of A respectively. Moreover, these values occurs when ~x is taken to be a corresponding unit eigenvector of the eigenvalue.

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