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One more EigenValues proof

  1. Jan 15, 2008 #1
    Question 1
    Let A be an nxn matrix such that [tex](A-I)^{2}=O[/tex] where [tex]O[/tex] is the zero matrix
    Prove that if [tex]{\lambda}[/tex] is an eigen value of A then [tex]{\lambda}=1[/tex]
    My attempt
    If [tex](A-I)^{2}=O[/tex] then [tex]A=I[/tex] (1)
    if [tex]{\lambda}[/tex] is an eigen value of A then [tex]Ax={\lambda}x[/tex] (2)
    replace (1) in (2) [tex]Ix={\lambda}x[/tex] , but [tex]Ix=x[/tex] therefore [tex]{\lambda}=1[/tex]

    Question 2
    If A is an orthogonal Matrix, then prove that det(A)=+-1
    My attempt
    if A is orthogonal then [tex]AA^{T}=I[/tex] and [tex]A^{-1}=A^{T}[/tex]
    therefore [tex]AA^{-1}=I[/tex] and [tex]det(AA^{-1})=1[/tex] . Where does the + - comes from?
     
  2. jcsd
  3. Jan 15, 2008 #2

    Dick

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    1) You can't say A=I. It's not necessarily true. You do know (A-I)(A-I)x=0. Now let x be an eigenvector...
    2) You've got det(A*A^T)=1. What can you say about the relationship between det(AB) and det(A) and det(B)? How about between det(A) and det(A^T)?
     
  4. Jan 16, 2008 #3
    Dick
    1)How do you know that (A-I)(A-I)x=0? Why not [tex](A-I)(A-I)x={\lambda}x[/tex].
    If (A-I)(A-I)x=0 and x being the eigen vector, this suggests that [tex]{\lambda}=0[/tex].

    2)I know that [tex]det(A^{T})=det(A) [/tex] but how does that prove that
    [tex]det(A)=+-1)[/tex] ?
     
  5. Jan 16, 2008 #4

    Dick

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    The problem says (A-I)^2=0. Operating on a vector means (A-I)(A-I)x=0. (A-I)x=Ax-Ix. If [tex]Ax={\lambda}x[/tex] that's [tex](\lambda-1)x[/tex]. Now apply the other (A-I). At the end you have a NUMBER times a nonzero vector equaling zero. So the number is zero.

    det(A*A^T)=1=det(A^T)*det(A)=det(A)^2. Solving that equation for det(A) is just like solving x^2=1 for x. What are the solutions?
     
    Last edited: Jan 16, 2008
  6. Jan 16, 2008 #5

    HallsofIvy

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    Dick's original point was simply that you cannot, from M2= 0, with M a matrix and 0 the zero matrix, conclude that M= 0 because there are "zero-divisors" in the algebra of matrices. For example,
    [tex]M= \left(\begin{array}{cc} 1 & 1 \\ -1 & -1 \end{array}\left)[/tex]
    has the property that M2= 0.

    Of course, saying M2= 0 means that M2x= 0x= 0 for any vector x. Separate that into M(Mx)= 0 and solve "twice".
     
  7. Jan 16, 2008 #6
    for the first question I came up with this: [tex]0=(A-I)^2=A^2-2A+I.[/tex] so if [tex]Ax=\lambda x,[/tex] for some
    vector [tex]x \neq 0,[/tex] then [tex]A^2x=\lambda^2x,[/tex] and [tex]0=(\lambda^2 - 2\lambda + 1)x=(\lambda - 1)^2x.[/tex] thus [tex]\lambda = 1.[/tex]
    This looks very right, doesn't it?

    For question 2, I figue that x^2=1 yields x=+-1.
    Thanks To all.
     
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