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Diagonalization 2

  1. May 17, 2009 #1
    I am reading yet another theorem and was wondering If I could get more clarification on it.

    Theorem 5.2: Let A be in [tex]M_n_x_n(F)[/tex]. Then a scalar [tex]\lambda[/tex] is an eigenvalue of A if and only if det(A - [tex]\lambda[/tex][tex]I_n[/tex]) = 0.

    Proof: A scalar [tex]\lambda[/tex] is an eigenvalue of A if and only if there exists a nonzero vector v in [tex]F^n[/tex] such that Av= [tex]\lambda[/tex]v, that is, (A - [tex]\lambda[/tex][tex]I_n[/tex])(v) = 0. By theorem 2.5, this is true if and only if A - [tex]\lambda[/tex][tex]I_n[/tex] is not invertible. However, this result is equivalent to the statement that det(A - [tex]\lambda[/tex][tex]I_n[/tex]) = 0.

    Theorem 2.5: Let V and W be vector spaces of equal (finite) dimension, and let T: V --> W be linear. Then the following are equivalent:
    (a.) T is one-to-one.
    (b.) T is onto.
    (c.) rank(T) = dim(V).

    Question: Could someone explain how the following sentence is true: "By theorem 2.5, this is true if and only if A - [tex]\lambda[/tex][tex]I_n[/tex] is not invertible."

    Thanks a lot,

  2. jcsd
  3. May 17, 2009 #2
    If it were invertible/nonsingular, then the only solution would be the zero vector. But by definition v is a nonzero vector.
  4. May 18, 2009 #3


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    I'm not sure why "theorem 2.5" is appended, this is a question about the sentence in theorem 5.2 only isn't it?

    In any case, the equation [itex]Av= \lambda v[/itex] is equivalent to [itex]Av- \lambda v= 0[/itex] which is equivalent to [itex](A- \lambda I_n)v= 0[/itex]. If [itex]A- \lambda I_n[/itex] were invertible, we could solve the the equation by multiplying both sides by that inverse: [itex]v= (A- \lambda I_n)^{-1})0[/itex] and, since any linear transformation of the 0 vector is the 0 vector, we must have v= 0, contradicting the fact that there was a non-zero vector satisfying [itex]Av= \lambda v[/itex].
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