Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Good afternoon.

I'm studying for my finals of this year, currently studying for the exam "Numerical Linear Algebra". I'm trying to solve some of the questions the teacher asked the past years (for preparation), but I can't seem to figure out a few, namely this one. We've got this setup:

  • $x=\Bigl[{-}\dfrac{1}{\sqrt{2}},{-}\dfrac{1}{\sqrt{2}},0\Bigr]$

    $Ax=[2,0,0]$

  • $x=\Bigl[{-}\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},0\Bigr]$

    $Ax=[0,-3,0]$

  • $x=[0,0,1]$

    $Ax=[0,0,6]$

The questions were as follows: 1. Determine the SVD of A (WITHOUT COMPUTING THE ACTUAL MATRIX) 2. Determine the condition number of A 3. Determine the rank and the determinant

Questions 2 and 3 are quite logical (condition number = largest SVD divided by the smallest, rank = number of non zero SV's and the determinant is the product of the SV's, if I'm not mistaken), but I can't seem to find a logical answer or explanation to question 1. So I'm hoping anyone here has a pointer in the right direction? I'm not asking for the asnwer (since I have to be able to replicate it on the exam, most likely), but some hints would be appreciated.

Thanks.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Since the three given vectors are orthonormal and their images are orthogonal, these are the right singular vectors. Thus, the singular value decomposition of $A$ is $$ A \;=\; \begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6\end{bmatrix}\begin{bmatrix}-1/\sqrt{2} & -1/\sqrt{2} & 0 \\ -1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 1\end{bmatrix}. $$ The rows of the last matrix are the given vectors, the diagonal entries of the middle matrix are the norms of the images of these vectors, and the columns of the first matrix are the normalizations of these images.

share|improve this answer
    
That makes insanely good sense, thanks. Only question remaining is: As far as I understand, the SV's have to be ordered from highest to lowest on the diagonal, or doesn't that matter? –  Timmy Nelen Jun 18 '11 at 17:49
    
There are different possible conventions. If you want to switch the order of the diagonal entries, all you have to do is switch the columns of the first matrix and the rows of the last matrix in the same way. –  Jim Belk Jun 18 '11 at 18:05
    
Alright, I've got it :) If I swap columns 1 & 3 in the first matrix, I swap the 2 and the 6 in matrix 2 (Thus making the diagonal [2,3,6]), and swap the first and third rows in the third matrix I have an SVD where the elements are in decreasing order (as per the standard that got utilized in our lectures), and thus I have a correct solution for it, thank you :) –  Timmy Nelen Jun 18 '11 at 18:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.