# Linear Algebra- Commuting matrix

## 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]

## 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:

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HallsofIvy
Homework Helper
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!

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.