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Math Help - Need to find basis for KerP

  1. #1
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    Need to find basis for KerP

    Hi guys! This is my first post so hopefully you can help me out. I also am not sure how to edit the text to make it look better so if anyone wants to help me with that, Id be glad to clean up the post.

    Here's the problem:
    Let Mnxn be the collection of all nxn matrices. Defined is a linear transform P:
    P: Mnxn --> Mnxn
    A |--> P(A) = (1/2)*(A - A^T)
    where A^T is the transpose of A.


    What I already did was find KerP which is shown below:
    KerP = {A: (1/2)*(A -A^T)=0} = {A = A^T}


    That's all fine and dandy but I cant figure out how to find the basis for KerP without some numbers involved. If I had a matrix A with numbers, I could do it but I guess I cant figure out how to generalize it. My guess is that there wouldnt be a basis but I wouldnt know how to go about proving that either. Thanks for any help. I'll be online for at least the next 8 hours so I can reply quickly as well.
    Last edited by Jiftim; April 17th 2010 at 12:57 PM.
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  2. #2
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    What does it mean if A=A^T?
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  3. #3
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    It would mean that P(A)=0 all the time for KerP which says to me that there is no basis. Perhaps my understanding of what a basis is is wrong though? The only other information I recall related to this is that KerP=NullP which is why I feel I would need numbers to solve this problem.
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  4. #4
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    You have a misstep with your ker.

    Ker(A)=(\mathbf{x}\in\mathbb{R}|A\mathbf{x}=0)
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    You have a misstep with your ker.

    Ker(A)=(\mathbf{x}\in\mathbb{R}|A\mathbf{x}=0)

    wouldnt that be Ker for some vector tranformation? As far as I understand, this linear transformation is different because it doesnt have a standard matrix A as you are implying.

    What I already did was find KerP which is shown below:
    KerP = {A: (1/2)*(A -A^T)=0} = {A = A^T}
    Otherwise I could have also written it as:
    KerP = {F in Mnxn: (1/2)*(F -F^T)=0} = {F = F^T}

    If that suits you better. (Also, this is how it was done in my class on another example so if it is wrong, I will need correction)
    Last edited by Jiftim; April 17th 2010 at 02:06 PM.
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  6. #6
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    \frac{1}{2}(A-A^T)\mathbf{x}=\mathbf{0}

    \begin{bmatrix}<br />
  a_{11}-a_{11}      & \cdots & a_{n1}-a_{1n}      \\<br />
  \vdots & \ddots & \vdots \\ <br />
  a_{1n}-a_{n1}      & \cdots & a_{nn}-a_{nn}<br />
\end{bmatrix}*\mathbf{x}=\mathbf{0}<br />

    This is what we essential have.

    Now, if A=A^t, then our matrix A=0
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  7. #7
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    That is true but that tells me there is no basis for KerP since, then, dimKerP=0 (or dimNulA=0 for the above) or am I still missing something else? I also feel as though my notation is horrible so feel free to correct me on it.
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  8. #8
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    Helpisplaying a formula - Wikipedia, the free encyclopedia

    LaTeX Online Equation Editor

    Also, if you double click the latex display in the forum, you will see how it is called.
    Last edited by dwsmith; April 19th 2010 at 02:13 PM.
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  9. #9
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    Ok but a basis for KerP isnt based off of x from my understanding. It would be based off of A. The question also asks me to find a basis for KerP which implies to me that there is a specific m number of vectors Im to look for. The next question also asks for rankP but I was just going to get that using the Rank Theorem. In this situation, I feel the question is begging for either a specific number or variable in which I can satisfy the Rank Theorem in the next question.
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  10. #10
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    The Ker can be 0
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  11. #11
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    hmm I feel as though I said that all along. Thanks anyway I guess. Still doesnt make a whole lot of sense though.
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  12. #12
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    Still dont get it

    If anyone else has input, further explanation would be helpful as I am still fuzzy on this topic.
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  13. #13
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    A little bit of clarity.

    Alright, so my professor went a little further into it today and we found dimKerP = (n+1)(n/2). She explained the basis to be a set of matrices rather than a set of vectors that span Mnxn which was very helpful.

    Since i didnt get it before, I still need to find rankP, so my question now is what is dimMnxn? I think it is n^2 but I'm not positive. Can someone help me with that or verify that what I think is right?
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  14. #14
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    Yes, dwsmith was thinking of the kernel of a matrix operation on vectors. Here, the "vectors" are themselves matrices and P(A)= (1/2)(A- A^T) is a transformation of matrices.

    The kernel is, as you say, all A such that A= A^T.

    Now, are you working with real matrices or are you allowing matrices with complex entries? If real then the kernel consists of symmetric matrices. For example, in 2 by 2 matrices of real numbers, the kernel consists of all matrices of the form \begin{bmatrix}a & b \\ b & c\end{bmatrix}= \begin{bmatrix}a & 0 \\ 0 & 0\end{bmatrix}+ \begin{bmatrix}0 & b \\ b & 0\end{bmatrix}+ \begin{bmatrix}0 & 0 \\ 0 & c\end{bmatrix}
    and a basis is
    \{\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}.

    In general n by n matrices, a basis can consists of all matrices have "1" at one entry on the diagonal, all others 0, and matrices where a_{nm}= a_{mn}= 0 for one pair (m,n), m\ne n, all other entries 0.

    For matrices over the complex numbers, you also have to take the complex conjugate so we would have to include matrices with a_{mn}= i, a_{nm}= -i.

    Since you teacher said there were n(n+1)/2 matrices in the basis, you must be working with real matrices.
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  15. #15
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    That is exactly what she was explaining to us! Thanks for that. I was having trouble figuring out how to put it onto the forum in a logical manner. (and yes we're working with real matrices)
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