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Norm of a Matrix

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\textbf{A}[/itex] be an m x n matrix and [itex]\lambda = \max\{ |a_{ij}| : 1 \leq i \leq m, 1 \leq j \leq n \}[/itex].

    Show that the norm of the matrix [itex]||\textbf{A}|| \leq \lambda \sqrt{mn}[/itex].

    2. Relevant equations
    The definition I have of the norm is that [itex]||\textbf{A}||[/itex] is the smallest number such that [itex]|\textbf{Ax}| \leq ||\textbf{A}|| \, |\textbf{x}|[/itex] for all [itex]\textbf{x}[/itex] in ℝn.


    3. The attempt at a solution
    I let [itex]\textbf{y} = \textbf{Ax} [/itex] and so [itex]|\textbf{y}| = \sqrt{\sum_{i=1}^{m}y_i^2}[/itex].
    I started by looking at a matrix [itex]\textbf{A}[/itex] with all the same entries so that any entry could be thought of as [itex]\lambda[/itex]. So when you calculate [itex]|\textbf{y}|[/itex] you will get [itex]m \lambda^2[/itex]'s under the radical along with sum of the [itex]x_1, \dots, x_n[/itex] squared. So in the more general case that not all the entires of [itex]\textbf{A}[/itex] are equal I can see how [itex]\lambda[/itex] would provide an upper bound.
    But I'm having trouble seeing how the [itex]\sqrt{n}[/itex] comes into it.
     
  2. jcsd
  3. Nov 17, 2013 #2
    So let A = ##\begin{pmatrix}
    a& a & a \\
    a & a & a\\
    \end{pmatrix}##

    (It's easier to type a than ##\lambda##)

    and multiply it by X = (x,y,z). You get AX = ##\begin{pmatrix}
    ax +& ay + & az \\
    ax+ & ay + & az\\
    \end{pmatrix}##

    You started correctly computing the ||AX|| and got to where |AX| = a##\sqrt 2 \sqrt{x^2 + y^2 + z^2}##.

    To see where the n comes in take X = (1,1,1,). Now what is ||AX|| less than?

    In finishing up, keep in mind that the < must work for every possible X. So it is perfectly possible there is an X which forces ||A|| lower. This is a bound, but for a given A it may not be the least upper bound.
     
  4. Nov 17, 2013 #3
    If X = (1,1,1) then [itex]|\textbf{Ax}| = a \sqrt{2}\sqrt{3}[/itex], but I guess I am missing what happens if X is a different vector, say (1,2,3). Then [itex]|\textbf{Ax}| = a \sqrt{2}\sqrt{14}[/itex], which isn't less than [itex] a \sqrt{2}\sqrt{3}[/itex]. I think in this case [itex]\textbf{||A||}= \frac{a\sqrt{72}}{\sqrt{14}}[/itex] so in this case [itex]\textbf{||A||} \, \textbf{|x|}< \lambda \sqrt{mn} [/itex]. But my trouble is generalizing it.
     
  5. Nov 17, 2013 #4
    After fiddling a little bit with the inequalities, it seems like in the general case it would be helpful to show that [itex]\frac{x_1 + x_2 + \dots + x_n}{\sqrt{x_1^2 + \dots + x_n^2}} < \sqrt{n}[/itex].
    Would this be a proper approach? And if so, is there some glaringly obvious fact I'm missing that would help me show this?
     
  6. Nov 17, 2013 #5
    Let X = ##(x_1,x_2,x_3)##. Let |x| be the largest of the 3 components. From our previous work we have ||AX|| ## \le a \sqrt 2 \sqrt {x_1^2+x_2^2 + x_3^2} \le a \sqrt 2 \sqrt 3 |x|##.

    Go back to your definition of ||AX||. This is the smallest number C such that ##||Ax|| \le C ||x||## for every vector X. We have shown (generalizing from the 3x2 case) that there is one vector x = (1,1,1...) such that ||Ax|| ## \le a \sqrt m \sqrt n |x| \le a \sqrt m \sqrt n||X||##. So there is no way that C can be greater than ##a \sqrt m \sqrt n##. It could be smaller than that, which is why we have the inequality.

    All this computation depended on A consisting of all a's. To finish up, you need to redo this deduction when A has varying entries ##a_{ij}##. In this case, choose ##a = max|a_{ij}|## and proceed more or less as we did before.
     
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