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Sketching Graphs using Eigenvalues

  1. Jun 14, 2013 #1
    1. The problem statement, all variables and given/known data
    For the conic, 5x2+4xy+5y2=9, find the direction of the principal axes, sketch the curve.

    I found the eigenvalues as
    3,7 but have no idea whether the 'new' equation is
    is there a way to determine which 'way' it goes?

    I took a guess and just continued using the first formula:
    I found the eigenvectors by substituting the eigenvalues and got:

    λ=3, V1 = (-1,1)
    λ=7, V2 = (1,1)

    I then thought the principal axes would therefore be:
    1/√2 (-1,1) and 1/√2 (1,1)
    yet the answer seems to indicate the principal axes as, (1,-1) and (1,1), why is that? I thought you had to normalise the vectors to find the principal axes..
  2. jcsd
  3. Jun 15, 2013 #2


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    x' and y' are just names. They don't relate in any particular way to x and y respectively. Might as well be u and v. So the two versions you offer are really the same.
    They're just directions. I don't see any requirement for them to be normalised.
  4. Jun 16, 2013 #3


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    "Axes" are lines not vectors- they do not have a specific length nor can they be written as vectors. The axes are neither "[itex](1/\sqrt{2})(-1, 1)[/itex]" and [itex](1/\sqrt{2})(1, 1)[/itex] nor "(-1, 1) and (1, 1)". The axes are lines in the directions of those vectors: y= -x in the direction of (-1, 1) and y= x in the direction of (1, 1).
  5. Jun 17, 2013 #4
    Okay, that seems to make more sense. So does order really not matter after we've found the eigenvalues?

    Also, I asked my lecturer this, but she couldn't properly explain the concept of diagonlisation using nullity.
    |2 1 0|
    |0 2 0|
    |0 0 -3|

    I'm trying to determine whether that is diagonizable or not, but have trouble understanding how to do it using the rank, nullity theorem.
  6. Jun 17, 2013 #5


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    You can't "using the rank, nullity theorem". You have to actually find the eigenvalues and then try to find the eigenvectors. Since the matrix here is diagonal, its eigenvalues are just 2 and -3, the numbers on the diagonal. There will be a one dimensional subspace of eigenvectors corresponding to eigenvalue 3 but we do not know yet if the eigenspace corresponding to eigenvalue 2 has dimension one or two. If it has dimension two, then there are two independent vectors in it and, adding an eigenvector corresponding to eigenvalue 3 gives a basis for R3 and the matrix, written in that basis, is diagonal. If it has dimension 1, the matrix cannot be diagonalized.
    If [itex]\begin{bmatrix}x \\ y \\ z \end{bmatrix}[/itex] is an eigenvalue corresponding to eigenvalue 2, then we must have
    [tex]\begin{bmatrix}2 & 2 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -3\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= 2\begin{bmatrix}x \\ y \\ z \end{bmatrix}[/tex]
    [tex]\begin{bmatrix}2x+ 2y \\ 2y \\ -3z \end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}[/tex]

    which is equivalent to the three equations 2x+ 2y= 2x, 2y= 2y, and -3z= 2z. The first equation says y= 0 and the third equation says z= 0. x is undetermined so eigenvectors corresponding to eigenvalue 2 are all multiples of [itex]\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}[/itex]. The eigenspace has dimension 1 so the matrix is NOT diagonalizable.
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