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Linear Algebra- Commuting matrix

  • Thread starter Roni1985
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  • #1
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Homework Statement


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]


Homework Equations





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:

Answers and Replies

  • #2
HallsofIvy
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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!
 
  • #3
201
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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!
This is some amazing explanation :)
Thank you very much.
This was very helpful.

Thanks a lot, HallsofIvy.
 

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