My Math Forum I can't solve this four question...need help plz

 Linear Algebra Linear Algebra Math Forum

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

Attached Images
 Untitled.png (26.8 KB, 369 views)

 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 $\sqrt{\frac{11}{2}}$
 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 rank-nullity theorem, $Dim(Ker A) + rank A= 4$. A vector $v$ is in $Ker A$ iff $Av= 0v = 0$. 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 $rank A= 3$. 2) For a random point on the first line $p_1= (\alpha , \ 1 + \alpha , \ 2)$ and for a random point on the second line $p_2= (1 + \beta , \ -1 + \beta , \ 3)$. The distance between the two lines is the minimum distance between random points $p_1$ and $p_2$ so that we're minimizing the distance function $d= \sqrt{ (\alpha - 1 - \beta)^2 + (2 + \alpha - \beta)^2 + 1} = \sqrt{2\alpha^2 + 2\beta^2 - 4\alpha \beta - 2\beta +2\alpha + 6}$. It's equivalent to minimizing $f= d^2 = 2\alpha^2 + 2\beta^2 - 4\alpha \beta - 2\beta +2\alpha + 6$ $\dfrac{\partial f}{\partial \alpha}= 4\alpha - 4\beta + 2$ and $\dfrac{\partial f}{\partial \beta}= 4\beta - 4\alpha - 2$. We want these partials to simultaneously equal 0 so that $\alpha - \beta= -\frac{1}{2}$. Our required distance is $d= \sqrt{ (\alpha - 1 - \beta)^2 + (2 + \alpha - \beta)^2 + 1} = \frac{\sqrt{22}}{2}$. 3)Let the center matrix be A so that $Q(x)= \langle Ax , x \rangle$. $Q'(x)(v) = lim_{t \rightarrow 0}\dfrac{Q(x + tv) - Q(x)}{t} = \langle x , Av \rangle + \langle Ax , v \rangle = 2\langle Ax , v \rangle$ because A is a symmetric matrix. As such, $Q'(x) = 2Ax$ and we will examine this derivative for the critical points of the functional $Q$. Observe that $Det A= 0$ 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 $B= (1 , \ -1 , \ 1 , \ -1)^T$. All critical points $x$ of Q satistfy $Q'(x) = 2Ax = 0$ 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 $Det(M - \lambda I)= 0$, we get eigenvalues of $\lambda _1= cos\theta + isin\theta$ and $\lambda _2= cos\theta - isin\theta$. 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:
 Originally Posted by AfroMike 2) For a random point on the first line $p_1= (\alpha , 1 + \alpha , 2)$ and for a random point on the second line $p_2= (1 + \beta , -1 + \beta , 3)$. The distance between the two lines is the minimum distance between random points $p_1$ and $p_2$ so that we're minimizing the distance function $d= \sqrt{ (\alpha - 1 - \beta)^2 + (2 + \alpha)^2 + 1} = \sqrt{2\alpha^2 + \beta^2 - 2\alpha \beta + 2\beta +2\alpha + 6}$.
Hi Mike, there's a small error in there. It should be:

$d= \sqrt{ (\alpha - 1 - \beta)^2 + (2 + \alpha - \beta)^2 + 1}$

$\alpha= -\frac{1}{2} \ and \ \beta = 0$

$d=\sqrt{\frac{11}{2}}$

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 $-2 , 0 , 2 , 4$ according to the arrangement of the column eigenvectors. If the eigenvalues were all non-negative, then we would have had a semi-positive definite matrix characterized by $\langle Ax , x \rangle \geq 0$ but we do not have this. This means we have no absolute (global) extrema for the functional $Q(x)= \langle Ax , x \rangle \geq 0$ . I hope you are clear with the argument that all critical points of $Q$ are scales of $B := (1 , \ -1 , \ 1 , \ -1)^T$ and you must test the two appropriate scales of B to see whether they are local minima or local maxima. $||B||= 2$, so you must test $x_1= \frac{1}{2}(1 , \ -1 , \ 1 , \ -1)^T$ and $x_2= -\frac{1}{2}(1 , \ -1 , \ 1 , \ -1)^T$. This requires examination of the Hessian matrix $Q''(x)$ which is precisely $2A$. 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 $Q$. However, for the minimum absolute value, we examine $|Q|$. Whenever $|Q|= 0$, we have a minimum absolute. Both $x_1$ and $x_2$ 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 $\mathbb{R}^4$ which is compact. Hence, at least one maximizer and at least one minimizer of $x \mapsto x^TQx$ 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?

 Tags plz, questionneed, solve

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post tayfur Calculus 8 June 18th, 2013 09:42 AM kozhin Number Theory 1 February 17th, 2013 02:42 PM darkangel777 Math Events 1 January 27th, 2010 02:28 PM mungbean_jones Abstract Algebra 0 December 31st, 1969 04:00 PM darocker077 Abstract Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top