Linear Algebra- Commuting matrix

[Roni1985]

Homework Statement

Determint the subspace fo R^2x2 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 I am a little off with the notations :\

 

[HallsofIvy]

Okay, so
A= \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}

Saying that B= \begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix} commutes with A means
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

\begin{bmatrix}b_{11} & b_{12} \\ -b_{21} & -b_{22}\end{bmatrix}= \begin{bmatrix}b_{11} & -b_{12} \\ b_{21} & -b_{22}\end{bmatrix}

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

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

With A= \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}
we have
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

\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}
So we must have b_{11}+ b_{21}= b_{11}, which means b_{21}= 0, b_{12}+ b_{22}= b_{11}+ b_{12}, which means b_{11}= b_{22}, and b_{22}= b_{21}+ b_{22}, which means b_{21}= 0.

That's exactly what you have! Very good!

 

[Roni1985]

Okay, so
A= \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}

Saying that B= \begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix} commutes with A means
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

\begin{bmatrix}b_{11} & b_{12} \\ -b_{21} & -b_{22}\end{bmatrix}= \begin{bmatrix}b_{11} & -b_{12} \\ b_{21} & -b_{22}\end{bmatrix}

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

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

With A= \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}
we have
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

\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}
So we must have b_{11}+ b_{21}= b_{11}, which means b_{21}= 0, b_{12}+ b_{22}= b_{11}+ b_{12}, which means b_{11}= b_{22}, and b_{22}= b_{21}+ b_{22}, which means b_{21}= 0.

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.