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Matrix Banach Space

  1. Jul 13, 2009 #1
    Let M_n(R) be the n x n matrices over the reals R. Define a norm || || on M_n(R) by ||A||= sum of absolute values of all the entries of A. Further define a new norm || ||* by ||A||* = sup{||AX||/||X||, ||X||!=0}.
    Show that

    1. M_n(R) under || ||* is complete.
    2. If ||A||<1, then I-A is nonsingular, where I is the identity matrix.
    3. The set of nonsingular matrices in M_n(R) is open.
    4. Find ||B||*, where B is 2x2 and b_11=1, b_12=2, b_21=3, b_22=4.

    There is a series of over 10 questions on the norm || ||. I've solved most of them but I've been stuck on (have no clue for) these ones above for a week.

    I'd appreciate any hints.
     
  2. jcsd
  3. Jul 13, 2009 #2
    Problem 1. Would completeness be easier in the other norm? Is there a relation between the norms?

    Problem 2. Once you prove completeness, then you can show that the series for [tex](I-A)^{-1}[/tex] converges.

    Problem 3. Use Problem 2.
     
  4. Jul 13, 2009 #3

    quasar987

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    For 4., notice that by linearity, if B' denotes the linear map associated to the matrix B, then ||B||*=sup{||B'(x)||: ||x||=1}. And R² can be identified (i.e. is isomorphic) as a normed linear space to C (via the correspondance (x,y)<-->x+iy). Under this correspondence, then, the map B' becomes the map C(x+iy)=(x + 2y) + i(3x + 4y) and so ||B||*==sup{||C(z)||: ||z||=1}. I have transported the problem from R² to C simply because in C, the unit vector are easily parametrized: they are just the e^it for t in [0,2pi]. And so the problem of finding ||B||* has been reduced to the simple calculus problem of finding max{||C(cos(t)+isin(t))||: t in [0,2pi]}.
     
  5. Jul 14, 2009 #4
    Would you know how to show that ||AB||*<=||A||*||B||*? This would be helpful for Problem 2.

    Also, how is ||B||* equal to sup{||B'(x)||: ||x||=1}? Notice that in my definition of || ||*, X refers to an nxn matrix.
     
    Last edited: Jul 14, 2009
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