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Derive an expression to find how many times an eigen value is repeated

  1. Feb 20, 2013 #1
    Hello !

    I have an upper triangular matrix for an operator T in which an eigen value has been repeated s times in total.

    Derive an expression for s .

    My thoughts : ( Let * imply contained in )
    then :I know that :

    (a)

    Null T0 * Null T1 *.....*Null Tdim V = Null Tdim V + 1 = ........

    (b) Will i have to investigate the effect of higher powers of ( T - k I ) where k is the intended eigen value ??

    (c) the book which i am reading : Sheldon Axler's Linear Algebra hasn't introduced Jordan form as of now.

    Any direction for this will be appreciated. Thanks
    Can i prove it from these results ?
     
    Last edited: Feb 20, 2013
  2. jcsd
  3. Feb 21, 2013 #2

    micromass

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    Re: Derive an expression to find how many times an eigen value is repe

    I have no idea what you're trying to find. What do you mean with "an expression"? Do you have to find some expression?? This is a very vague question...
     
  4. Feb 21, 2013 #3

    micromass

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    Re: Derive an expression to find how many times an eigen value is repe

    The only thing I can think of is that the eigenvalue comes up s times on the diagonal. Maybe they mean that?
     
  5. Feb 21, 2013 #4
    Re: Derive an expression to find how many times an eigen value is repe

    The answer given states that s = dim [ Null ( T - k I )dim V ]
    where k is the corresponding eigen value.
     
  6. Feb 21, 2013 #5
    Re: Derive an expression to find how many times an eigen value is repe

    An expression means in this context a formula to find number of times an eigen value is repeated in an upper triangular matrix.
     
  7. Feb 21, 2013 #6

    micromass

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    Re: Derive an expression to find how many times an eigen value is repe

    So you need to prove that formula for s? How is this thread different from your previous thread then?
     
  8. Feb 21, 2013 #7
    Re: Derive an expression to find how many times an eigen value is repe

    I thought i ended up confusing a lot of things over there in that thread. So, i wrote afresh, This is what i meant actually.
     
  9. Feb 21, 2013 #8

    micromass

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    Re: Derive an expression to find how many times an eigen value is repe

    And why does Theorem 8.10 not answer the question for you?? I think it basically says and proves what you want.
     
  10. Feb 21, 2013 #9
    Re: Derive an expression to find how many times an eigen value is repe

    Hi

    I found proof by induction unconvincing ; It assumes that the result is already true. ( It does not give an intuition .. )

    I really want to derive the expression considering a situation , say, when i never knew what the answer is going to be in which case, probably induction is not going to work.

    I have thought about it and i think the answer may lie in investigating the behavior of higher powers of ( T - k I ) but i seem to be stuck for more than a week now, which is frustrating :(
     
  11. Mar 13, 2013 #10
    Found it

    ok guys, i have finally found a proof. Took me long .I will post it for common good :)

    if an eigen value λ is repeated r times on the main diagonal of M(T) [ where M(T) denotes the matrix associated with the linear mapping T ] , then M(T - λ I ) has r zeroes on the main diagonal.

    Speaking of higher powers of M(T - λ I ) ( say kth power ) , notice that under any circumstance, the non zero diagonal element of M(T - λ I ) would simply be their kth power for M(T - λ I ). -------------------- (A)

    since, the eigen vectors for non distinct eigen values may/may not be linearly independent

    => dim [ null (T - λ I ) ] ≤ r
    => dim [ range (T - λ I ) ] ≥ n-r

    Now, we know that :

    null (T - λ I )0 [itex]\subset[/itex] null (T - λ I )1[itex]\subset[/itex] ......... [itex]\subset[/itex] null (T - λ I )m
    =null (T - λ I )m+1=...... = null (T - λ I )dim V =..

    => 0 < dim null (T - λ I ) < ... < dim null (T - λ I )m = dim[ null (T - λ I )m+1 ] = .... = dim null (T - λ I )dim V = ....

    ...................... (1)


    We also know that range (T - λ I )0 [itex]\supset[/itex] range (T - λ I )1[itex]\supset[/itex] .......... [itex]\supset[/itex] range (T - λ I )m = range (T - λ I )m+1 =... = range (T - λ I )dim V = ...


    => n > dim range (T - λ I )1 > ..... > dim range (T - λ I )m = dim range (T - λ I )m+1=...... dim range (T - λ I )dim V

    ........................ (2)

    after carefully analysing the statement (A) , it states that the minimum dimension of range of any power of ( T - λ I ) = n-r .

    If we try to look at the safest boundary conditions :

    max [ dim range (T - λ I ) ] = n-1

    We already know that max [ dim [range (T - λ I )m ] ] = n-r and not less than that .

    => maximum value of m from statement (2) = r ---------------- (3)
    => dim range (T - λ I )r = n-r
    => dim null (T - λ I )r = r

    => from (1) : dim null (T - λ I )dim V = r .

    Hence, there you have the expression for the algebraic multiplicity of an eigen value.
     
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