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Eigenvalue and Eigenvectors

  1. Dec 3, 2016 #1
    1.
    1) Given 2x2 matrix A with A^t = A. How many linearly independent eigenvectors is A?
    2) Is a square matrix with zero eigenvalue invertible?

    2; When it comes to whether it is invertible; the det(A-λ* I) v = 0
    where det (A-λ * I) v = 0 where λ = 0
    We get Av = 0, where the eigenvector is zero. But this proves anything?
    Or it is possible to prove that det(A) ≠ 0? Im not sure about how to prove it

    I feel I lack some theory behind it all.
    Im completely blank on the first, so thanks for all your help
     
    Last edited by a moderator: Dec 3, 2016
  2. jcsd
  3. Dec 3, 2016 #2

    hilbert2

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    Suppose ##v## is an eigenvector of matrix ##A## with eigenvalue zero: ##Av=0##. If there's an inverse matrix ##A^{-1}##, we should have ##A^{-1}Av=A^{-1}0##, or equivalently ##v=A^{-1}0##. Is this possible? What properties are required from an eigenvector?
     
  4. Dec 3, 2016 #3

    vela

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    It doesn't really make sense to write, as you did, ##\det(A-\lambda I)\vec{v} = 0##. An eigenvector satisfies ##A\vec{v} = \lambda \vec{v}##, which you can write equivalently as ##(A-I\lambda)\vec{v} = 0##. For this system to have a non-trivial solution, you must have ##\det(A-\lambda I) = 0##.

    You might find it useful to review the basics about matrices, inverses, and determinants. If you don't know what implies what, you're really not in a position to try prove anything.

    Try explicitly solving for the eigenvectors of a symmetric 2x2 matrix. That is, start with ##A = \begin{pmatrix} a & b \\ b & d\end{pmatrix}## and find the eigenvalues and eigenvectors. See if that gets you anywhere.
     
  5. Dec 3, 2016 #4
    Hi mr-feeno,

    Invertible means the same thing as in "invertible function". We are seeking a "reverse" function that can give us back the original vector from the image of a vector in the range of the matrix transformation. That is, consider the matrix A as a function f(v) = Av, where Av is ordinary matrix multiplication of the matrix A with an arbitrary vector v. If Av = 0v = 0 for any particular vector v, notice what happens if we try any vector that is a scalar multiple of v as an input to our function. Suppose w = kv for any scalar k. Then f(w) = Aw = Akv = kAv = k*0 = 0. So every single vector that is a scalar multiple of v will be sent to 0 if v is sent to 0. Think of trying to invert this action. Given 0, we would like the inverse function to tell us which vector was sent to 0. Is it possible, given that more than one vector was sent to 0?
    If you need to do it with arithmetic, then we recall that a square matrix is not invertible if and only if its determinant is 0 (this theorem is part of a large laundry list of basic interrelated determinant properties that you should have somewhere in your text or notes). As you have shown that the determinant is 0, you may then imply that the square matrix is not invertible (also referred to as being "singular"). The reason why was described in greater detail above.
    For part 1., the fact that it is a 2x2 matrix that is symmetric means it must be of the form
    [tex]\begin{bmatrix}a & b\\ b & c\end{bmatrix}[/tex]
    for some numbers a, b, and c. Try to find the eigenvalues and eigenvectors of this type of matrix. What does this tell you?
     
  6. Dec 3, 2016 #5
    [tex] v=A^-1 *0[/tex] By definition, No.It must be non-trivial. Can you also think ;If 0 is an eigenvalue, then that matrix would be similar to a matrix whose determinant is equal to zero?
     
  7. Dec 3, 2016 #6
    I'll do it. Thanks for a a detailed answer
     
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