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Homework Help: Infinity/One Norm Multiplication

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data

    This isn't actually a homework question but i thought this would be the right place for it... I am doing exam review and this question is giving me difficulties.

    Consider the 3x3 diagonal[1,3,1] matrix A. Find nonzero vectors x in ℝ[itex]^{3}[/itex] such that [itex]||Ax||_{3} = ||A||_{3} * ||x||_{3}[/itex]

    Do the same, but with one norms.

    3. The attempt at a solution

    So my initial guess is that every value in x has to be the same for the comparison to work. I have to write vector x as a general vector i assume... So would x = {a1,a2,a3} where a1=a2=a3 suffice?

    For one norms it seems to depend on whether values in x are positive or negative.
  2. jcsd
  3. Nov 17, 2013 #2
    Where you have to start is with definitions. If x = (##x_1,x_2,x_3##) what is ||x||##_3## in terms of its components? And what is ||A||##_3## in terms of its components. If you will write out those definitions we can proceed to the next step.

    I don't think your guess about the vector is right. We're going to have to get to the eigenvectors of A, but that is not step 1.
  4. Nov 18, 2013 #3
    I think there was a formatting error with the original post, i need to find the infinity norm, not the 3-norm.

    The infinity norm of a vector is the max absolute value in the vector. the infinity norm of a matrix is its max absolute row sum.

    If we take x=(0,1,2) then we see that ||Ax|| is different from ||A|| * ||x|| ... The max row sum of the matrix is 7, versus 5 * 2 = 10... It seems that it only works if x contains all the same elements.
  5. Nov 18, 2013 #4
    Well, the ##\infty## norm is different.

    However, the max absolute row sum of A is 3. Where did you get 7?
    If you have the right ||A|| the problem is not difficult.
  6. Nov 18, 2013 #5
    The max abs row sum of Ax is 7 since it would be 0*0 + 1*1 + 2*3. The max absolute row sum of A would be 5 since the matrix is

    3 1 0
    1 3 1
    0 1 3

    and therefore the middle row has the highest value

    however this doesn't answer the question really, i need to come up with a general solution. What this is supposed to show is that ||Ax|| =/= ||A|| ||x|| when the contents of x are not all the same.

    I would prefer to learn how to do this kind of a question without guessing and coming up with cases which would make the proposition true... Is there not some sort of method for solving it?
  7. Nov 18, 2013 #6
    The 3x3 diagonal matrix [1,3,1] is

    1& 0 & 0 \\
    0 & 3 & 0\\
    0 & 0 & 1
    \end{pmatrix} ##

    A diagonal matrix has only entries on the diagonal and zero's everywhere else.

    So where did you get

    3 1 0
    1 3 1
    0 1 3

    Is the matrix not diagonal? Did you present the problem to me properly?
  8. Nov 19, 2013 #7
    I took the definition of the diagonal matrix verbatim from my old midterm (there is even a diagram of the matrix)... Perhaps my prof has a different definition for diagonal matrices? In his matrix there are only two zeroes in the top right and bottom left corners.
  9. Nov 19, 2013 #8
    Definitely not diagonal. I'll work on the problem you were given, but later today.
  10. Nov 19, 2013 #9
    It looks like your original guess is correct, that the only vectors which work are multiples of (1,1,1) (there's nothing like working with the right matrix). I'll start you off on showing this, as I'm not supposed to do the whole problem for you.

    We have A = ##\begin{pmatrix}

    If ||A|| is the largest row sum we have ||A|| = 5. We are to find a vector X = (a,b,c) such that
    ||AX|| = 5||X||.

    We know that AX =##\begin{pmatrix}
    3a + b\\
    a + 3b + c\\
    b + 3c

    Suppose ||X|| = a. Then we want ||AX|| = 5a, which means
    3a + b = 5a or a + 3b + c = 5a or b + 3c = 5a.

    The first equation gives b = 2a, which contradicts that ||X|| = a.

    The second & third equations give
    3b + c = 4a
    b + 3c = 5a

    Solve these two equations for b and c in terms of a (just the usual 2 equations in 2 unknowns). This solution should give you another contradiction allowing you to conclude that ||X|| ≠ a (if we want an X that works).

    If you try to take ||X|| = c, you will get the same result because the situation is symmetric. That leaves ||X|| = b. Set up your 3 equations again, and you can show that they give a = b = c, but that no other solution is available.

    Re the word "diagonal" really there is no alternate definition, so I'm inclined to think your prof just had a typo. He could have said A is symmetric, which at least is correct, although I couldn't have deciphered the 1,3,1 into which symmetric matrix he wanted.
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