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Check if a scalar is an eigenvalue of a matrix

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data
    We have a matrix Anxn (different than the identity matrix I) and a scalar λ=1. We want to check if λ is an eigenvalue of A.

    2. Relevant equations
    As we know, in order for λ to be an eigenvalue of A, there has to be a non-zero vector v, such that Avv

    3. The attempt at a solution
    Avv
    Av=1v
    Av=v
    A=I

    But we know that A is different than I, so λ is not an eigenvalue of A.
    Is my attempt right?

    Thanks in advance for your assistance.
     
    Last edited: Jun 13, 2012
  2. jcsd
  3. Jun 13, 2012 #2

    Ray Vickson

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    Your last equation is nonsense. From Av = v you cannot conclude that A = I. All you can conclude is that v must be a solution of the homogeneous linear system (A-I)v = 0.

    RGV
     
  4. Jun 13, 2012 #3

    HallsofIvy

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    In particular, the only operations defined on a vector space are addition of vectors and multiplication or division by a scalar. You cannot "divide both sides" by a vector.

    (If Av=Bv for all vectors, v, then you can conclude that A= B. But not if Av= Bv for some vector, v.)
     
  5. Jun 13, 2012 #4
    In other words, you can't divide both sides of the equation by the vector [itex]v[/itex]. Linear operators like [itex]\underline A[/itex] may be written down using matrices and said to operate on vectors through matrix multiplication, but what's really going on is that [itex]A[/itex] is a linear function, which is why trying to factor [itex]v[/itex] from both sides is nonsense.

    Eigenvalues satisfy the characteristic equation that [itex]\det (\underline A - \lambda \underline I) = 0[/itex].

    (PS. In particular, a linear operator acting on a vector has a general form [itex]\underline A(v) = (v\cdot e_1)a + (v \cdot e_2)b + (v \cdot e_3)c + \ldots{}[/itex] where [itex]a,b,c[/itex] are vectors. Unless you know something about the structure of the operator, dividing both sides by a vector doesn't give you anything useful.)
     
  6. Jun 13, 2012 #5
    Thank you all for your replies.
    One last question:
    If we think of this theoretically, in the equation Av=v, there has to be a matrix A that will be multiplied by v and the result will be the same vector v. I understand that we cannot divide both parts of the equation by v, but could you please give me an example with a non-identity matrix A and a non-zero vector v that satisfy the equation Av=v?
     
  7. Jun 13, 2012 #6
    Sure. Consider a linear operator on [itex]\mathbb R^4[/itex].

    [tex]\underline A(e_1) = e_1 \\
    \underline A(e_2) = -e_2 \\
    \underline A(e_3) = e_3 \\
    \underline A(e_4) = -e_4[/tex]

    Any linear combination of [itex]e_1, e_3[/itex] is an eigenvector of this operator with eigenvalue 1.
     
  8. Jun 13, 2012 #7
    Thank you!
     
  9. Jun 13, 2012 #8

    Mark44

    Staff: Mentor

    $$A = \begin{bmatrix}1 & 1\\ 0 & 1 \end{bmatrix}$$
    $$\lambda = 1, v = \begin{bmatrix}1 \\ 0 \end{bmatrix}$$
     
    Last edited: Jun 13, 2012
  10. Jun 13, 2012 #9
    Thanks!
     
  11. Jun 15, 2012 #10
    If (A - I)v = 0, what condition needs to be satisfied in order for v not to have the trivial solution v = 0?

    Chet
     
  12. Jun 15, 2012 #11

    Mark44

    Staff: Mentor

    If A - I is invertible, the only solution is v = 0.
    If A - I does not have an inverse, then v = 0 is still a solution, but there are also nonzero solutions.

    So in both cases, v = 0 is a solution. The only difference is whether that is the unique solution or it is one of an infinite number of solutions.

    You can tell whether a square matrix is invertible -- its determinant is nonzero. If an inverse does not exist, the determinant is zero.
     
  13. Jun 15, 2012 #12
    Thanks Mark. That was the response I was trying to elicit from the OP.

    Chet
     
  14. Jun 16, 2012 #13

    HallsofIvy

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    Yes, but you phrased it poorly- by leaving out a single word! You said, "If (A - I)v = 0, what condition needs to be satisfied in order for v not to have the trivial solution v = 0?"

    v= 0 is, as Mark44 said, always a solution. What you meant to say was, "If (A - I)v = 0, what condition needs to be satisfied in order for v not to have only the trivial solution v = 0?"

    Notice, by the way, that the "eigenvalue" question is "existence and uniqueness" turned on its head. A solution, the trivial solution, always exists for the equation [itex]Av=\lambda v[/itex]. [itex]\lambda[/itex] is an eigenvalue if and only if that solution is NOT unique.
     
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