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Homework Help: Matrix multiplication and the dot product

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that A is an n x n matrix such that A(Transpose)A=I. Let x be any vector in R^n. Show that llAxll=llxll; that is, multiplication of x by A produces a vector Ax having the same length as x.


    2. Relevant equations

    Sqrt(x(transpose)x)=llxll

    3. The attempt at a solution

    So I started setting up the equation wth an n x n matrix...

    l a b l =A
    l c d l

    l a c l= A (transpose)
    l b d l

    A(transpose)A= I=
    l a^(2) +c^(2) ab+cd l
    l ba+dc b^(2)+d^(2)l

    then I let the vector x=

    l x1 l
    l x2 l


    then from there I was going to use the distance formula on Ax

    but I wasn't sure if doing the dot product of Ax is the same thing as matrix multiplication of Ax, and my second problem is that I am not sure how to take the square root of a 2x2 matrix if they are the same thing.

    If someone can answer those to questions for me I think I can finish the problem.

    Thank you.
     
  2. jcsd
  3. Jan 25, 2010 #2

    Dick

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    You don't have to work with any explicit matrices. If you are writing expressions like Ax then you should be thinking of x as a column vector. So ||x||=sqrt(transpose(x)x). So ||Ax||=sqrt(transpose(Ax)Ax). Can you express transpose(Ax) in terms of transpose(A) and transpose(x)?
     
  4. Jan 26, 2010 #3
    I'm not too sure how to do that.

    Would this argument work?

    If A(transpose)A=I, then A=A(transpose)=I.

    If A=I, then Ax=Ix=x

    therefore llAxll=llIxll=llxl
     
  5. Jan 26, 2010 #4

    vela

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    No, because your first step is wrong. There are matrices other than I where [itex]A^TA=I[/itex].

    Try looking up the properties of transpose, in particular, the transpose of a product.
     
  6. Jan 26, 2010 #5
    ok I looked up all I know about transpose, and I also looked up all I know about I...
    so here is what I know, but I ended up using other theorums to solve it.

    Thrm 10: If A and B are m x n matrices and c is an n x p matrix, then:

    1. (A+B)^T=A^T +B^T
    2.(AC)^T=C^T * A^T
    3. (A^T)^T=A


    What I know about I...
    .I is a matrix whose diagnal is all 1's and the rest are 0's.
    .AI=A
    .I is nonsingular (the only soultion is the trivial solution)
    . (A^(-1)*A)^T=I


    Here is my second attempt. I ended up using mainly other thrms then the ones above but I am pretty sure I have it this time. Thank you to eveyone who has helped me thus far, and anyone who spends the time to read my proof.


    If I is non singular then A and A transpose must be nonsingular,because if either A or A transpose was singular then I would have to be singular and I happens to be nonsingular. The lemma states, "Let P,Q, and R be n x n matrices such that PQ=R. IF either P or Q is singular then so is R.

    Then I also know that all non singular matrices can row reduce to I. If all non singular matrices can reduce to I, and A is non singular then I must be equal to A, by thrm 1.

    thrm 1 states, " if one of the 3 row operations is applied to a system of linear equations, then the resulting system is equivalent to the original system.

    So because any matrix time I is equal to that matrix and I is equal to A then...

    A=I,

    so Ax=Ix=x, therefore

    llAxll=llxll
     
  7. Jan 26, 2010 #6

    vela

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    This is completely wrong. Just because a matrix is non-singular doesn't mean it equals I. It can be reduced to I, but the reduced matrix isn't equal to the original matrix.

    An obvious counterexample to your logic: Take A=2I. It's obviously non-singular, but |Ax| = 2|x|. "Aha!" you say, "but [itex]A^TA \ne I[/itex]!" Yes, that's true, but your proof doesn't rely on that assumption either, so according to your proof's logic, A=2I should preserve the length of vectors.

    So look at property 2 of theorem 10 and use the hint Dick gave above.
     
  8. Jan 26, 2010 #7
    oh wow that made things much easier but I am still a little confused.

    so I have (Ax)^T=x^(T)*A^(T)

    llxll=llAxll

    sqrt(x^(T)*x)=sqrt((x^(T)*A^(T))(Ax))


    x^(T)*x=(x^(T)*A^(T))(Ax)

    My problem starts here now. I thought the order of multiplication mattered, and for me to get this to simplifiy I would have to break that order up I feel like because I would multiply the A^T by the A first rather than the B^T by the A^T. Does order matter here? And if it does, does it disable me from being able to do A^T by A first?
     
  9. Jan 26, 2010 #8

    vela

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    What you're thinking of is the fact that matrix multiplication is not commutative, that is, [itex]AB\ne BA[/itex] generally, but it is associative, so [itex](AB)C = A(BC)[/itex].
     
  10. Jan 26, 2010 #9

    Dick

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    You've got (x^(T)*A^(T))*(A*x). You are confusing commutativity with associativity. It's true you can't change the order of matrices but you can regroup them. I.e. (AB)(CD)=A(BC)D. That's just associativity. And it is true.
     
  11. Jan 26, 2010 #10
    ahhhhhhhh got ya. Thank you so much for all the help everyone.
     
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