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October 15th, 2013, 04:48 AM  #1 
Newbie Joined: Oct 2013 Posts: 4 Thanks: 0  I can't solve this four question...need help plz
Please help..

October 15th, 2013, 05:23 AM  #2 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: I can't solve this four question...need help plz
Here a geometric answer to number 2: First move the origin, so that the first line passes through (0,0,0) and the second (1, 2, 1). This preserves the distance between the lines. Note that the direction vector in each case is the line y = x. So, the normal to the first line at the origin is y = x. The second line has equation x = 1 + u, y = 2 +u, z = 1. So, y = x goes directly under this line at (1.5, 1.5, 0). So, the nearest point to the origin on the second line is (1.5, 1.5, 1). So, the distance between the lines is 
October 15th, 2013, 09:53 AM  #3 
Senior Member Joined: Dec 2012 Posts: 372 Thanks: 2  Re: I can't solve this four question...need help plz
Have you just posted your homework? Well, never mind  here are solutions, and I hope you understand. 1.) By ranknullity theorem, . A vector is in iff . Since the eigenvalues are distinct, their algebraic multiplicities and geometric multiplicities match. In particular, the geometric multiplicity of 0 equals the dimension of the kernel of A, which is 1. It follows that . 2) For a random point on the first line and for a random point on the second line . The distance between the two lines is the minimum distance between random points and so that we're minimizing the distance function . It's equivalent to minimizing and . We want these partials to simultaneously equal 0 so that . Our required distance is . 3)Let the center matrix be A so that . because A is a symmetric matrix. As such, and we will examine this derivative for the critical points of the functional . Observe that so that the dimension of the kernel of A is at least one. From the given hint of the computed eigenvectors, we are able to detect the kernel of A to be spanned by the vector . All critical points of Q satistfy so that they must be in the kernel of A. Since in addition, we are working with vectors of unit length, you need only to focus on appropriate scales of the vector B. I'll leave it up to you to detect which is the maximizer, minimizer and so on. 4.) Solving , we get eigenvalues of and . I'll leave it up to you to verify and complete this problem. 
October 15th, 2013, 10:24 AM  #4  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: I can't solve this four question...need help plz Quote:
Which leads to Which leads to Which is what I got by the geometric method!  
October 15th, 2013, 10:36 AM  #5 
Senior Member Joined: Dec 2012 Posts: 372 Thanks: 2  Re: I can't solve this four question...need help plz
I concur Pero, and have accordingly corrected my post. Thanx

October 17th, 2013, 09:18 AM  #6 
Newbie Joined: Oct 2013 Posts: 4 Thanks: 0  Re: I can't solve this four question...need help plz
Thank you for your help. But I still do not know how to solve question 3, and I do not understand the meaning of the Hint... could you please explain?

October 18th, 2013, 08:07 AM  #7 
Senior Member Joined: Dec 2012 Posts: 372 Thanks: 2  Re: I can't solve this four question...need help plz
Ok. You were given the eigenvectors so that you can obtain the eigenvalues more easily and hence detect the category of the center matrix. You can quickly infer that the eigenvalues of this matrix are respectively according to the arrangement of the column eigenvectors. If the eigenvalues were all nonnegative, then we would have had a semipositive definite matrix characterized by but we do not have this. This means we have no absolute (global) extrema for the functional . I hope you are clear with the argument that all critical points of are scales of and you must test the two appropriate scales of B to see whether they are local minima or local maxima. , so you must test and . This requires examination of the Hessian matrix which is precisely . But the Hessian always has both positive and negative eigenvalues meaning that all critcal points are saddle points. In conclusion, no local or global extrema exist for . However, for the minimum absolute value, we examine . Whenever , we have a minimum absolute. Both and as described above satisfy all criteria for being minimum absolutes under the given constraints. 
November 20th, 2013, 10:11 AM  #8 
Senior Member Joined: Dec 2012 Posts: 372 Thanks: 2  Re: I can't solve this four question...need help plz
wasd3123, in retrospect I have made a blunder in my attempt to problem 3. I hope it didn't cost you much. Your given functional which is continuous has its domain to be the unit sphere in which is compact. Hence, at least one maximizer and at least one minimizer of exist on the domain. The reason using classical calculus is a flop, is that the unit sphere has an empty interior; and so it cannot accomodate classical calculus. There are other optimization techniques that are valid here. Did you consult further on this problem? What did you gather?


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