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Finding the 2-norm of a 2x2 matrix

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


    2. Relevant equations

    [itex]\left(
    \begin{array}{cc}
    4 & 2 \\
    2 & 1 \\
    \end{array}
    \right)[/itex]

    3. The attempt at a solution

    There are definitely different ways to solve this. I will use Lagrange multipliers.

    [itex]\textbf{A}x =
    \left(
    \begin{array}{cc}
    4x_1+2x_2 \\
    2x_1+x_2 \\
    \end{array}
    \right)[/itex]

    [itex](\left|\textbf{A}\textbf{x}\right|)^2=(4x_1+2x_2)(4x_1+2x_2)+(2x_1+x_2)(2x_1+x_2)=18x_1^2+20x_1 x_2+5x_2^2[/itex]

    Use Lagrange multipliers at this step, with the condition that the norm of the vector we are using is x. The goal is to find the unit vector such that A maximizes its scaling factor.

    [itex]f = 18x^2+20x_1x_2+5x_2^2+\mu(x_1^2+x_2^2)[/itex]

    Find the derivatives in the ::x_1:: and ::x_2:: directions and set each to 0.

    [itex]\frac{\partial f}{\partial x_1}=36x_1+20x_2+2\mu x_1=0[/itex]
    [itex]\frac{\partial f}{\partial x_2}=20x_1+10x_2+2\mu x_2=0[/itex]

    I am not sure where to go from here. I know that the norm of the matrix is 5, and I tried comparing coefficients to get different values for ::\mu:: and pick the right ::\mu:: for the right unit vector. If I were to directly compare coefficients, I get two different values for ::\mu::.
    First value:
    [itex]\mu = -18[/itex]
    Second value:
    [itex]\mu = -5[/itex]

    These are associated with two unit vectors which would make the the two partial derivative equations hold true:
    (1,0)
    and
    (0,1)

    These are definitely wrong, because if I use the first unit vector (1,0), the associated magnitude of Ax would be root 18. If I were to use the second unit vector (0,1), the associated magnitude would be root 5.

    What am I doing wrong?
     
  2. jcsd
  3. Nov 17, 2013 #2
    I know this has to be the right method, but the last step that I am taking is wrong. Does anyone have any insight into what is going on? I'm wasting hours on this thing.
     
  4. Nov 18, 2013 #3
    I guess that's a no.

    I take it I can just add up the squares of the indices, take the square root of the whole thing, and get the answer for the norm, which is 5.

    But, why doesn't this work for the following matrix?

    [itex]\left(
    \begin{array}{ccc}
    -1 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 4 \\
    \end{array}
    \right)[/itex]

    The norm of this matrix is 4, but, when I add the squares, I get [itex]\sqrt{21}[/itex].
     
  5. Nov 18, 2013 #4
    I guess my question's not that interesting. Oh well. The test is in a few minutes. Too late now.
     
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