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Homework Help: Linear Algebra- Commuting matrix

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Determint the subspace fo R2x2 consisting of all matrices that commute with the given matrix:
    a) A=[(1,0)^T,(0,-1)^T]

    c) A=[(1,0)^T,(1,1)^T]


    2. Relevant equations



    3. The attempt at a solution
    What I need to show is that AB=BA. So, I am trying to see when it happens...

    I think I got them right buy let me see:
    let B be any matrix in the subspace

    a) S= {B | b12=b21=0} OR it should be S= {B | b12=-b21}

    c) S= {B | b21=0 and b11=b22 }

    are they correct ?

    perhaps Im a little off with the notations :\
     
    Last edited: Mar 31, 2010
  2. jcsd
  3. Mar 31, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Okay, so
    [tex]A= \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}[/tex]

    Saying that [tex]B= \begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}[/tex] commutes with A means
    [tex]AB= \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}= \begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}= BA[/tex]

    [tex]\begin{bmatrix}b_{11} & b_{12} \\ -b_{21} & -b_{22}\end{bmatrix}= \begin{bmatrix}b_{11} & -b_{12} \\ b_{21} & -b_{22}\end{bmatrix}[/tex]

    So the conditions are that [itex]b_{12}= -b_{12}[/itex] which gives [itex]b_{12}= 0[/itex] and [itex]-b_{21}= b_{21}[/itex] so [itex]b_{21}= 0[/itex].

    Your first statement is correct- this is the set of diagonal matrices.

    With [tex]A= \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}[/tex]
    we have
    [tex]AB= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}\begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}= \begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}= BA[/tex]

    [tex]\begin{bmatrix}b_{11}+ b_{21} & b_{12}+ b_{22} \\ b_{21} & b_{22}\end{bmatrix}= \begin{bmatrix}b_{11} & b_{11}+ b_{12} \\ b_{21} & b_{21}+ b_{22}\end{bmatrix}[/tex]
    So we must have [itex]b_{11}+ b_{21}= b_{11}[/itex], which means [itex]b_{21}= 0[/itex], [itex]b_{12}+ b_{22}= b_{11}+ b_{12}[/itex], which means [itex]b_{11}= b_{22}[/itex], and [itex]b_{22}= b_{21}+ b_{22}[/itex], which means [itex]b_{21}= 0[/itex].

    That's exactly what you have! Very good!
     
  4. Mar 31, 2010 #3
    This is some amazing explanation :)
    Thank you very much.
    This was very helpful.

    Thanks a lot, HallsofIvy.
     
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