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Math Help - Matrix of a Nilpotent Operator Proof

  1. #1
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    Matrix of a Nilpotent Operator Proof

    I'm trying to understand this one proof about the matrix of a nilpotent operator and I'm stuck on this one part. Here is the theorem I'm trying to prove and how the proof goes:

    Suppose N is a nilpotent operator on a vector space V. Then there is a basis of V with respect to which the matrix of N has the form:

    <br />
\[<br />
 \begin{bmatrix}<br />
  0 &  & * \\<br />
   & \ddots &  \\<br />
  0 &  & 0<br />
 \end{bmatrix}<br />
\]

    where all entries on and under the diagonal are 0 and the asterisk indicates the other entries.


    The proof starts by creating a basis for V. First a basis for null(N) is chosen and then extended to a basis of null(N^2). Then our basis is extended to a basis for null(N^3) and so on until null(N^{\dim{V}})=V. Now we have a basis of V.

    The text then says:
    "Now letís think about the matrix of N with respect to this basis. The
    first column, and perhaps additional columns at the beginning, consists
    of all 0ís because the corresponding basis vectors are in null(N). The
    next set of columns comes from basis vectors in null(N^2). Applying N
    to any such vector, we get a vector in null(N); in other words, we get a
    vector that is a linear combination of the previous basis vectors. Thus
    all nonzero entries in these columns must lie above the diagonal. The
    next set of columns come from basis vectors in null(N^3). Applying N
    to any such vector, we get a vector in null(N^2); in other words, we get a
    vector that is a linear combination of the previous basis vectors. Thus,
    once again, all nonzero entries in these columns must lie above the
    diagonal. Continue in this fashion to complete the proof."

    I understand how the entries in the first set of columns are all 0. However, I do not understand how the second set of columns are determined.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Anthonny View Post
    I'm trying to understand this one proof about the matrix of a nilpotent operator and I'm stuck on this one part. Here is the theorem I'm trying to prove and how the proof goes:

    Suppose N is a nilpotent operator on a vector space V. Then there is a basis of V with respect to which the matrix of N has the form:

    <br />
\[<br />
 \begin{bmatrix}<br />
  0 &  & * \\<br />
   & \ddots &  \\<br />
  0 &  & 0<br />
 \end{bmatrix}<br />
\]

    where all entries on and under the diagonal are 0 and the asterisk indicates the other entries.


    The proof starts by creating a basis for V. First a basis for null(N) is chosen and then extended to a basis of null(N^2). Then our basis is extended to a basis for null(N^3) and so on until null(N^{\dim{V}})=V. Now we have a basis of V.

    The text then says:
    "Now let’s think about the matrix of N with respect to this basis. The
    first column, and perhaps additional columns at the beginning, consists
    of all 0’s because the corresponding basis vectors are in null(N). The
    next set of columns comes from basis vectors in null(N^2). Applying N
    to any such vector, we get a vector in null(N); in other words, we get a
    vector that is a linear combination of the previous basis vectors. Thus
    all nonzero entries in these columns must lie above the diagonal. The
    next set of columns come from basis vectors in null(N^3). Applying N
    to any such vector, we get a vector in null(N^2); in other words, we get a
    vector that is a linear combination of the previous basis vectors. Thus,
    once again, all nonzero entries in these columns must lie above the
    diagonal. Continue in this fashion to complete the proof."

    I understand how the entries in the first set of columns are all 0. However, I do not understand how the second set of columns are determined.
    Think about it like this. Let (x_1,\cdots,x_m) be the selected basis for \ker N and \left\{x_1,\cdots,x_n\} the full basis constructed. Then, as the author said one has that for each element x_j of basis vectors one has that N(x_j)\in \ker N so that there exists \alpha_1,\cdots,\alpha_m\in F such that \displaystyle N(x_j)=\sum_{r=}^{m}\alpha_r x_r\quad\mathbf{(1)} and so in particular if \beta_1,\cdots,\beta_n we form the x_j^{\text{th}} column by listing them as the corresponding coefficients of \displaystyle \sum_{r=1}^{n}\beta_r x_r...but since the representation with respect to basis elements is unique we may conclude by \mathbf{(1)} that \beta_r=0 for r>m. But, since j>m this implies that the first element which is necessarily zero occurs at or above the basis. Make sense?
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  3. #3
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    ok, remember we have picked a basis B = {v1,...,vk} for V. so we can write v in V as: v = a1v1 + a2v2 +....+ akvk.

    hence N(v) = a1N(v1) + a2N(v2) + a3N(v3) +...+ akN(vk)

    now the columns of N will be the images N(vj), in the basis B. for illustration suppose dim(null(N)) = 3.

    relative to the basis B, v1,v2 and v3 have coordinates (1,0,0,...,0), (0,1,0,...,0) and (0,0,1,...,0). clearly N(v1) and N(v2) and N(v3) are all 0-vectors

    (in the basis B!). now, in the basis B, v4 has coordinates, (0,0,0,1,...,0). N(v4) is not 0, so the 4th column for N cannot be all 0's.

    but N^2(v4) = 0. this means that N(v4) is in null(N), so N(v4) = b1v1 + b2v2 + b3v3. so the B-coordinates of N(v4) are (b1,b2,b3,0,....,0).

    again, for the sake of illustration, suppose that v5 isn't in null(N^2), but is in null(N^3).

    then N(v5) is in null(N^2) so N(v5) = c1v1 + c2v2 + c3v3 + c4v4, so the B-coordinates of N(v5) are (c1,c2,c3,c4,....,0)

    (remember {v1,v2,v3,v4} form a basis for null(N^2)).

    do you see it now?
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  4. #4
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    The jth column of the matrix for N are the coordinates of N(\mathbf{e}_j) with respect to the constructed basis for V.

    Suppose \mathbf{e}_j \in \text{null}N^2. Then N(\mathbf{e}_j)\in \text{null} N. Hence N(\mathbf{e}_j) can be written as a linear combination of the basis for \text{null} N. This implies that the coordinates wrt to the rest of the basis for V are zero. That is, everything on and below the diagonal of the matrix for N in the jth column will be zero. You then continue arguing like this for columns corresponding to basis elements in \text{null} N^3, \text{null} N^4, etc.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Think about it like this. Let (x_1,\cdots,x_m) be the selected basis for \ker N and \left\{x_1,\cdots,x_n\} the full basis constructed. Then, as the author said one has that for each element x_j of basis vectors one has that N(x_j)\in \ker N so that there exists \alpha_1,\cdots,\alpha_m\in F such that \displaystyle N(x_j)=\sum_{r=}^{m}\alpha_r x_r\quad\mathbf{(1)} and so in particular if \beta_1,\cdots,\beta_n we form the x_j^{\text{th}} column by listing them as the corresponding coefficients of \displaystyle \sum_{r=1}^{n}\beta_r x_r...but since the representation with respect to basis elements is unique we may conclude by \mathbf{(1)} that \beta_r=0 for r>m. But, since j>m this implies that the first element which is necessarily zero occurs at or above the basis. Make sense?

    Yes, I completely understand how the proof works.
    Thank you!

    as well as the other replies which also made sense to me.
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