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A Matrix with Orthonormal Columns

  1. Jul 27, 2008 #1
    The problem statement, all variables and given/known data
    Let M be an n x n matrix. Prove that the columns of M form an orthonormal set if and only if M-1 = MT.

    The attempt at a solution
    Let's consider the following 2 x 2 matrix

    [tex]\begin{pmatrix} a & b \\ c & d \end{pmatrix}[/tex]

    If the inverse equals its transpose, i.e.

    [tex]\frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} a & c \\ b & d \end{pmatrix}[/tex]

    then

    [tex]\begin{align*}
    a & = \frac{d}{ad - bc} \\
    b & = \frac{-c}{ad - bc} \\
    c & = \frac{-b}{ad - bc} \\
    d & = \frac{a}{ad - bc}
    \end{align*}[/tex]

    or rather that ad - bc = 1. Does this mean that ab + cd = 0? Not necessarily: Consider a = 2 and b = c = d = 1 so that ad - bc = 2 - 1 = 1 but ab + cd = 2 + 1 = 3. What gives?

    And doeas ab + cd = 0 imply the equations above for a, b, c and d? I think not.

    I'm stumped.
     
  2. jcsd
  3. Jul 27, 2008 #2
    Just multiply out the matrices and think.
     
  4. Jul 27, 2008 #3
    [tex]\left| ad - bc \right|=1[/tex], but ad - bc may not necessarily be 1. For example, the columns of [tex]\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}[/tex] are orthonormal, but [tex]det \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}=-1.[/tex] I guess it would be easier to work from the relation [tex]\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}[/tex], which follows from the hypothesis that the transpose of M is its inverse. The equations should be simpler to work with. In general, the product of any 2 x 2 matrix M and its transpose [tex]M^T[/tex] may be expressed as : [tex]\begin{pmatrix} a & b \\ c & d \end{pmatrix} \times \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} (r_1,r_1) & (r_1,r_2) \\ (r_2,r_1) & (r_2,r_2) \end{pmatrix}[/tex], where [tex]r_i[/tex] denotes the ith row of M, and [tex](r_i,r_j)[/tex] denotes the scalar product of [tex]r_i[/tex] and [tex]r_j[/tex]. Now equate the latter expression with the identity matrix. This shows that the rows of M are orthonormal, and the argument can be extended to any (n x n) matrix. To show that the columns are also orthonormal, we could use the fact that if [tex]MM^T = I[/tex], then [tex] M^TM=I[/tex], and thence express the product [tex]M^TM[/tex] as we did above.
     
  5. Jul 27, 2008 #4

    tiny-tim

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    Homework Helper

    Hi e(ho0n3! :smile:

    Why make it so complicated? :rolleyes:

    In problems like this, just write out the definition!
    Hint: MMT = I means, for any i and j, ∑aikajk = Iij.

    Put i = j: then … ?

    Put i ≠ j: then … ? :smile:
     
  6. Jul 27, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I don't understand your point with this example. The inverse of
    [tex]A= \left[\begin{array}{cc}2 & 1 \\ 1 & 1\end{array}\right][/tex]
    is
    [tex]A^{-1}= \left[\begin{array}{cc}2 & -1 \\ -1 & 1\end{array}\right][/tex]
    NOT the transpose of A and so has nothing to do with this problem.
     
  7. Jul 27, 2008 #6
    OK. Looks like I really messed up on this one. Thank you all for the pointers.
     
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