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Mathematical proof

  1. Aug 22, 2010 #1
    I am looking for a rather simple proof that if the matrix A has eigenvalues>0, then (x^t)Ax>0 for any vector x not 0.

    My first tought was if the eigenvalues are bigger than 0, then (x^t)Ax=(x^t)"eigenvaulue"x="eigenvalue"(x^t)x>0, if x is nonzero and eigenvalues is bigger than 0.

    Is this proof good enough? I am a little unsure, becuase I think I have not proven it for all x, just the eigenvectors.
     
  2. jcsd
  3. Aug 22, 2010 #2

    HallsofIvy

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    I'm not sure what you mean by '(x^t)"eigenvaulue" x'. which eigenvalue?

    It is true that if A is diagonalizable, then it is similar to a diagonal matrix having the eigenvalues on the diagonal.

    For example, if A itself is
    [tex]A= \begin{bmatrix}a_1 & 0 \\ 0 & a_2\end{bmatrix}[/tex]
    where [itex]a_1[/itex] and [itex]a_2[/itex] are both positive, we have
    [tex]x^*Ax= \begin{bmatrix}x_1 & x_2 \end{bmatrix}\begin{bmatrix}a_1 & 0 \\ 0 & a_2\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix}[/tex]
    [tex]= a_1x_1^2+ a_2x_2^2[/itex]
    which is postive because it is the sum of two positive numbers (or one positive number and 0).

    But not all matrices are diagonalizable.
     
  4. Aug 22, 2010 #3
    You'll have a hard time proving this statement because it's not true.

    Consider the following matrix:

    [tex]
    A= \begin{bmatrix}9 & 5.5 \\ 1 & 1\end{bmatrix}[/tex]

    The eigenvalues of A are both positive (verify).

    Let x = (1,-3.1)T

    Then
    [tex] x^*Ax= \begin{bmatrix} 1 & -3.1 \end{bmatrix}\begin{bmatrix}9 & 5.5 \\ 1 & 1\end{bmatrix} \begin{bmatrix}1 \\ -3.1\end{bmatrix} = -1.54
    [/tex]

    The relationship you describe is true a) if A is symmetric or b) if (A+A*)/2 is positive definite.
     
  5. Aug 23, 2010 #4

    HallsofIvy

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    Well done.
     
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