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A linear operator T on a finite-dimensional vector space

  1. May 16, 2009 #1
    Definitions: A linear operator T on a finite-dimensional vector space V is called diagonalizable if there is an ordered basis B for V such that [tex][T]_B[/tex] is a diagonal matrix. A square matrix A is called diagonalizable if [tex]L_A[/tex] is diagonalizable.

    We want to determine when a linear operator T on a finite-dimensional vector space V is diagonalizable and, if so, how to obtain an ordered basis B = [tex]{v_1, v_2, ... , v_n}[/tex] for V such that [tex][T]_B[/tex] is a diagonal matrix. Note that, if D = [tex][T]_B[/tex] is a diagonal matrix, then for each vector [tex]v_j[/tex] in B, we have

    [tex]T(v_j)[/tex] = [SUMMATION: from i = 1 to n][tex]D_i_jv_i[/tex] = [tex]D_j_jv_j[/tex] = [tex](lambda_j)v_j[/tex]
    where (lambda_j) = Djj.

    Questions: Could someone explain the following:
    1. [tex]T(v_j)[/tex] = [SUMMATION: from i = 1 to n][tex]D_i_jv_i[/tex]

    2. And maybe touch upon the other two equality relation in the line above.



    Thanks,


    JL
     
  2. jcsd
  3. May 16, 2009 #2

    jbunniii

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    Re: Diagonalization

    T is a linear operator mapping from V to V. So for each element [tex]v_j[/tex] of the basis, you can represent [tex]T(v_j)[/tex] uniquely as a linear combination of [tex]\{v_1,\ldots,v_n\}[/tex]. The coefficients of that linear combination are precisely the elements of the j'th column of [tex][T]_B[/tex].

    [tex]T(v_j) = \sum_{i=1}^{n} D_{ij} v_i[/tex]

    expresses exactly what I wrote above: on the right hand side we are holding j (the column index) fixed, and summing over i (the row index).

    Now, if the matrix is diagonal, only one of the [tex]D_{ij}[/tex]'s in a given column is nonzero, which is why the next equality is true:

    [tex]\sum_{i=1}^{n} D_{ij} v_i = D_{jj}v_j[/tex]

    So now you have the fact that

    [tex]T(v_j) = D_{jj} v_j[/tex]

    Since [tex]v_j \neq 0[/tex] (because it is an element of a basis), this equality says precisely that [tex]D_{jj}[/tex] is an eigenvalue of T, and [tex]v_j[/tex] is a corresponding eigenvector. Thus the suggestive [tex]\lambda_j[/tex] in place of [tex]D_{jj}[/tex] in the last equality.
     
  4. May 16, 2009 #3

    HallsofIvy

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    Re: Diagonalization

    A linear operator, L, (and its associated matrix) is diagonalizable if and only if there is a basis for the space consisting entirely of eigenvectors of L. In that case we can 'diagonalize' L by D= P-1LP where P is the matrix having the eigenvectors of L as columns and D is the diagonal matrix having the eigenvalues of L on its main diagonal.
     
  5. May 16, 2009 #4
    Re: Diagonalization

    Thanks for the clarification. Now I have to try to commit this to memory.

    JL
     
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