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Eigenvalues And Eigenvectors Problems

  1. Apr 10, 2007 #1
    1. How to show (prove) the Cayley-Hamilton theorem :
    “Every matrix is a zero of its characteristic polynomial , Pa(A)=0”.
    2. A and B are n-square matrices, show that AB and BA have the same eigenvalues.
    3. Show that to say that “ 0is an eigenvalue of linear mapping U” is equivalent to “ U is singular”.
    4. Compare eigenvalues of U and eigenvalues of U[upperscript]-1[/upperscript].
     
  2. jcsd
  3. Apr 10, 2007 #2

    mathwonk

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    if you know about determinants,i thinkyou can see AB and BA have the same characteristic polynomial. If U stretches a vector by 2, what does Uinverse do to that vector?

    If you even know the definition of the words you can show that 0 beiung an eigenvalue is the same as being singular.

    the cayley hamilton theorem is tricky but the shortest proof i know is the lagrange determinant formula that implies that if A is a square matrix and adA is itsclassical adjoint, i.e. transpose of matrix of alternating subdeterminants,

    then [X-A][ad(X-A] = det(X-A).Id. (Cramers rule).

    Then since X-A divides the characteristic polynomial in the ring of matrices with polynomial entries, it also does so in the ring of polynomials with matrix coefficients, so by the non commutative root factor theorem, A is a root.
     
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