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Linear algebra - Transformation using a Matrix

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    let [itex]T:R^3 \rightarrow R^3[/itex]
    be a Linear map,
    and let [itex]B=\left \{ (1,1,1),(1,1,0),(1,0,1) \right \}[/itex]
    be a Basis

    and [itex](1,0,0)\in kerT[/itex]


    [itex][T]_B=\begin{pmatrix}
    a & 0 & b\\
    3 & 2a & 1\\
    2c& b & a
    \end{pmatrix}[/itex]

    a. find a,b,c
    b. find a Basis for ImT

    2. Relevant equations



    3. The attempt at a solution

    as for a.

    I think that what we need to do is find a general vector (x,y,z)
    and express (1,0,0) thorugh it
    if we multiply the matrix by what we got we will have to get (0,0,0) (because (1,0,0) is in the kernel)
    but I'm not sure how to express (1,0,0).
    I think the vector (1,0,0) in the basis B is -1(1,1,1)+1(1,1,0)+1(1,0,1) and so is
    (-x+y+z,0,0)

    and for the matrix we have
    -a+0+b=0
    -3+2a+1=0
    -2c+b+a=0

    from the second equation:
    a= 1
    from the first
    b=1
    from the third
    c=1

    Am I right?
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2

    lanedance

    User Avatar
    Homework Helper

    yeah look good calling the natural basis E, (1,0,0) in the B basis is as follows
    [tex](1,0,0)^T_E = (-1,1,1)^T_B [/tex]

    And as a check in E
    [tex]-1\textbf{b}_1+1\textbf{b}_2+1\textbf{b}_3 = -1(1,1,1)^T+1(1,1,0)^T+1(1,0,1)^T=(-1+1+1,-1+1+0,-1+0+1)^T =(1,0,0)^T [/tex]
     
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