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Math Help - diagonisable matrrices question

  1. #1
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    diagonisable matrrices question

    here is nxn matrices . \rho(A)=1 and tr(A)=n
    find the representative polinomial of A and tell if A is diagonisable.

    there is a way which my prof said is wronggg:
    i said that this A could be a representation matrices for some transformation
    and if \rho(A)=1 then dim ImT=1 too,so dim KerT=n-1

    so if we take a basis sfor this kernel w_1..w_{n-1} and expand it
    to w_1..w_{n}.

    so the first n-1 columns are zero columns exsept the last n'th column
    and from the law that the traces of similar matrices are equal i get
    that the diagonal of the representative matrices is all zero except there is 'n' in the corner
    and from here its very easy to prove that its diagonisable.

    but my prof said that its wrong because even if the representative matrice is diagonisable
    it doesnt say anything about A


    this is how my prof recommended to start to solve it:
    if \rho(A)=1 then we have eigen value 0 with n-1 eigen vectors.
    dont know how to continue from here

    ??
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  2. #2
    Super Member girdav's Avatar
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    Re: diagonisable matrrices question

    The way you showed is wrong because with your hypothesis we may have A=I.
    Maybe there is a typo in the hint given by your prof, since for example if A=I, 0 is not an eigenvalue.
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    Re: diagonisable matrrices question

    A is singular becausee its nxn matrices with ro=1 so det|A|=0
    so A cannot be I
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    Super Member girdav's Avatar
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    Re: diagonisable matrrices question

    If \rho(A) is not \max\left\{|\lambda |,\,\lambda \mbox{ eigenvalue of }A\right\}, can you define it?
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    Re: diagonisable matrrices question

    i cant understand your meaning
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    Super Member girdav's Avatar
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    Re: diagonisable matrrices question

    I only want to know what \rho(A) is if it's not a standard notation.
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    Re: diagonisable matrrices question

    ahhh ro(A) means the number of rows which differ zero after row reduction
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    Re: diagonisable matrrices question

    i believe transgalactic means: ρ(A) = rank(A).

    rank(A) = 1 --> nullity(A) = n-1.

    the nullity of A is the dimension of the null space of A, which is also the eigenspace corresponding to the eigenvalue 0.

    this tells us the geometric multiplicity of 0 is n-1, so the algebraic multiplicity of 0 is at least n-1.

    this, in turn, tells us that x^{n-1} |p(x) where p(x) is the characteristic polynomial of A.

    now A has to have at least 1 non-zero eigenvalue since nullity(A) < n. for the time being, let's call this eigenvalue λ, and let v be an eigenvector corresponding to λ.

    then if \{w_1,w_2,\dots,w_{n-1}\} is a basis for null(A), then

    \{w_1,w_2,\dots,w_{n-1},v\} is an eigenbasis for V, so A is diagonalizable.

    if P is the matrix whose columns are the eigenbasis vectors, then PAP^{-1} is the matrix with all 0's, and λ in the lower-right hand corner.

    and as you pointed out tr(A) = tr(PAP^{-1}) = n \rightarrow n = \lambda.

    so the characteristic polynomial is: x^{n-1}(x-n).
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    Re: diagonisable matrrices question

    "now A has to have at least 1 non-zero eigenvalue since nullity(A) < n"
    why is that?

    before you say
    " is an eigenbasis for V, so A is diagonalizable"

    you must first prive that this basis consists only from eigen vectors

    you havent prove that "v" eigen vector in this step
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  10. #10
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    Re: diagonisable matrrices question

    if nullity(A) = n, then A is the 0-matrix. the 0-matrix (and only the 0-matrix) has rank 0.

    A does not have rank 0, so A is not the 0-matrix. rank(A) = 1, so every vector in the column space of A,

    that is every vector in the image A(V), is a multiple of just one vector in V. let's call this basis of im(A), w.

    so for EVERY vector v in V, A(v) = cw, for some scalar c (which scalar depends on which vector v we choose).

    but w is in V, so A(w) = cw. thus w is an eigenvector for A, with eigenvalue c. because w is a BASIS vector, it is non-zero.

    can the eigenvalue c be 0? no. because if A(w) = 0w = 0, then every vector in the image of A would have to be 0 as well

    (w is a basis for im(A)), contradicting the fact that A is NOT the 0-matrix.

    so, A has an eigenvector with a non-zero eigenvalue (we can in fact, use any non-zero column of A as this eigenvector,

    and we know A HAS a non-zero column, because A is not the 0-matrix).

    in fact, A has a LOT of eigenvectors with non-zero eigenvalue, but they are all scalar multiples of each other. that is what rank(A) = 1 means.

    for example, suppose A is the matrix:

    [3 0 0]
    [1 0 0]
    [0 0 0]. then (3,1,0) is an eigenvector for A with eigenvalue 3. indeed, A(x,y,z) = (3x,x,0) = x(3,1,0),

    so {(3,1,0)} is a basis for the image of A. we get the same thing if A =

    [2 2 0]
    [1 1 0]
    [0 0 0]. here A(x,y,z) = (2x+2y,x+y,0) = (x+y)(2,1,0). either non-zero column works as an eigenvector.

    to find the eigenvalue, just find A(2,1,0) = (2+1)(2,1,0) = 3(2,1,0), so the eigenvalue is 3.
    Last edited by Deveno; June 28th 2011 at 06:35 PM.
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