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Basis of Kernel (matrices)

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Let V = M2(R) be the vector space over R of 2×2 real matrices. We consider the mapping
    F : V −> V defined for all matrix M belonging to V , by F(M) = AM +MA^T where A^T denotes the transpose matrix of the matrix A given below

    A =
    
    1 2
    −1 0
    
    Question is: Determine a basis of Ker(F)

    3. The attempt at a solution
    So I showed that F is a linear operator, and preserves scalar addition and multiplication.
    However I am lost as to how I can solve the equation:
    AM +MA^T = 0

    Any help appreciated, thanks :)
     
  2. jcsd
  3. Mar 5, 2008 #2
    AM + MA^T = 0, so AM = -MA^T. You know what A is and M is a 2x2 matrix, so pick some entres a, b, c, d for M and solve for them.
     
  4. Mar 5, 2008 #3
    What is zero in [itex]M_2(\mathbb{R})[/itex]? Just apply the definition of kernel using [itex]M = \left( \begin{array}{cc}
    m_{11} & m_{12}\\m_{21} & m_{22}\\\end{array} \right)[/itex] and you will find the basis.
     
    Last edited: Mar 5, 2008
  5. Mar 6, 2008 #4
    I get dimension of Ker = 4,

    What I did is write out the matrix and multiply it out (since we know A and M I took
    as a,b,c,d). After multiplying and adding, I get a system of 4 equations, (a,b,c)
    and solve them via Gauss to find how many are independent.

    Is this ok?
     
  6. Mar 6, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You solve the matrix equation by doing the work to write it out as separate equations for the components.
    Let
    [tex]M= \left(\begin{array}{cc}a & b \\ c & d\end{array}\right)[/tex]
    Then
    [tex]F(M)= AM+ MA^T= \left(\begin{array}{cc} 1 & 2 \\ -1 & 0\end{array}\right)\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)+ \left(\begin{array}{cc}a & b \\ c & d\end{array}\right)\left(\begin{array}{cc}1 & -1 \\ 2 & 0\end{array}\right)[/tex]
    [tex]= \left(\begin{array}{cc}a+2c & b+2d \\ -a & -b+d\end{array}\right)+ \left(\begin{array}{cc}a+ 2b & -a \\ c+2d & -c\end{array}\right)[/tex]
    [tex]= \left(\begin{array}{cc}2a+2b+2c & -a+b+2d \\ -a+ c+ 2d & -b- c+ d\end{array}\right)[/tex]
    For M be "in the kernel", that must be the 0 vector. Solve 2a+ 2b+ 2c= 0, -a+ b+ 2d= 0, -a+ c+ 2d= 0, and -b- c+ d= 0. If those equations are all independent, of course, the only solution will be a= b= c= d= 0, the 0 matrix. If not, then the kernel may have dimension 1, 2, 3, or 4. (Well, it's pretty obvious the dimension is not 4.)
     
  7. Mar 6, 2008 #6
    EDIT: Oh wait found a mistake in my math
     
  8. Mar 6, 2008 #7
    Yes did exactly that, by the way last equation should be -b-c

    So after this I solve the matrix of coefficients to see how many independent columns
    I have:

    2 2 2 0
    -1 1 0 2
    -1 0 1 2
    0 -1 -1 0

    and simplified matrix is:

    1 1 1 0
    0 1 2 2
    0 0 1 2
    0 0 0 4



    I get 4 independent columns once I solve via Gauss. So Ker F = {0}.. so is this
    dim 0?
     
    Last edited: Mar 6, 2008
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