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Eigenvalue Proof Help

  1. May 4, 2012 #1
    Hey everyone,

    I have a problem with over thinking things quite often, so I once again need help haha.
    How would you go about proving this:

    λ=0 is the only eigenvalue of A [itex]\Rightarrow[/itex] Ax=0 [itex]\forall[/itex]x

    Any help would be appreciated!
  2. jcsd
  3. May 4, 2012 #2
    Are you sure it's true? The matrix
    A = \left [\begin{array}{ccc}
    0 & 1 & 1 \\
    0 & 0 & 1 \\
    0 & 0 & 0 \\
    \end{array} \right ]
    has [itex] \lambda = 0 [/itex] with multiplicity 3. Multiplying [itex] A [/itex] by the ones vector does not, however, yield the zero vector.

    Perhaps you left out some constraints on [itex]A[/itex]?
  4. May 4, 2012 #3


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    Hey AsprngMathGuy and welcome to the forums.

    The result can be shown from the definition of the eigenvalue problem where you have:

    [itex]Ax = λx[/itex] which implies [itex](Ax - Iλx) = 0[/itex]

    If all your λ's are zero then the above reduces to [itex]Ax - 0*Ix = 0[/itex] which gives us [itex]Ax=0[/itex]. This only uses the definition of the eigenvalue problem and you simply plug in the value for λ to get your equation.
  5. May 4, 2012 #4


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    That is not correct.

    You have to change your matrix to this:

    A = \left [\begin{array}{ccc}
    -λ & 1 & 1 \\
    0 & -λ & 1 \\
    0 & 0 & -λ \\
    \end{array} \right ]
    [/tex] when you factor in the -λI term.

    It still gives you zero (like you said above), but substituting in λ in your equation will give you Ax = 0.

    You have to remember that you are solving an eigenvalue problem, and in doing this you want to usually assume your A is a non-singular matrix. In your example your A is a singular matrix and this will be useless.

    The idea with an eigenvalue problem is that you want to find what x gets 'scaled' in a linearly dependent way from your matrix A, which will help you decompose it into the eigenvectors by analyzing where the scaling happens.
  6. May 4, 2012 #5


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    Perhaps Chiro is misunderstanding the question. A, given by robertsj, is, in fact, an example of matrix having only A as eigenvalue but such that Ax is not always 0.

    IF there exist a basis for the vector space consisting of all eigenvectors (a "complete set of eigenvectors" so that A is diagonalizale) then Ax= 0 for all x. But in general, there may not exist such a set of eigenvectors. There will be some subspace such that Ax= 0 for all x in that subspace. In robersj's example, that would be the subspace of vectors <1, 0, 0>.
  7. May 4, 2012 #6


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    True HallsofIvy. Can't see the point for doing an eigendecomposition when you have no eigenvectors or can't find them.
  8. May 4, 2012 #7
    chiro: you're absolutely right if A were to be invertible, but that would be an unspecified constraint on A. As for assuming a nonsingular matrix, I work all the time with a method dealing with singular operators.
  9. May 4, 2012 #8


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    What kind of problems? Just curious.
  10. May 4, 2012 #9
  11. May 4, 2012 #10
    Thank you so much for your input. I actually misread the problem slightly. At the beginning it states to prove OR disprove the problem. So of course my mind goes right to attempting to prove it haha. Thanks!
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