Commutative 2x2 Matrices: Finding Solutions for AB = BA

[RJLiberator]

Homework Statement

Let A =
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}

Find all 2 x 2 matrices B such that AB = BA.

Homework Equations

http://euclid.colorado.edu/~roymd/m3130/Exam2sol.pdf

The Attempt at a Solution

I let B =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} and set AB=BA.

From here I see that a and d must be 0, and b=c must be true.

So the answer will be that all matrices that are commutative will be of form:

\begin{bmatrix}
0 & b \\
b & 0
\end{bmatrix}

And there is no other possible commutative matrix outside of this form.

1. Is this correct?
2. Is there any further proof of this needed?

Thank you kindly.

 

[pasmith]

Homework Statement

Let A =
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}

Find all 2 x 2 matrices B such that AB = BA.

Homework Equations

http://euclid.colorado.edu/~roymd/m3130/Exam2sol.pdf

The Attempt at a Solution

I let B =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} and set AB=BA.

From here I see that a and d must be 0, and b=c must be true.

So the answer will be that all matrices that are commutative will be of form:

\begin{bmatrix}
0 & b \\
b & 0
\end{bmatrix}

That must be wrong, because the 2x2 identity matrix commutes with every 2x2 matrix but is not of that form.

What did you actually get for AB and BA? I would suggest double-checking those calculations.

And there is no other possible commutative matrix outside of this form.

1. Is this correct?

You have the right idea, but have not executed it correctly.

 

[RJLiberator]

So to get this right:

if A = \begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}

All possible commutative matrices with matrix A should be in the form:

\begin{bmatrix}
0 & b \\
b & 0
\end{bmatrix}This is the wrong answer? It seems to be right when I calculate it. I get the same answer either way. AB = BA

 

[pasmith]

$\begin{pmatrix} 0 & b \\ b & 0 \end{pmatrix}$ is a subset of the matrices you are looking for. It can't be all of them, because the identity $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ commutes with every 2x2 matrix.

Recheck your initial calculations with $B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

 

[RJLiberator]

Ahhh, I see.

I calculated it out and found that while b=c, also a=d:

\begin{bmatrix}
a & b \\
b & a
\end{bmatrix}

This makes sense to me as the original answer was a subset of this.

Is there any further proof needed to show that is all?

 

[SammyS]

All possible commutative matrices with matrix A should be in the form:

[0bb0]​

\begin{bmatrix} 0 & b \\ b & 0 \end{bmatrix}This is the wrong answer? It seems to be right when I calculate it. I get the same answer either way. AB = BA

To follow up on what pasmith is asking you to do:

What do you get for the product $\displaystyle \ \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} \ \ \ ?$

What do you get for the product $\displaystyle \ \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \ \ \ ?$(I see you posted you answer just before I posted this.)

That looks good.

 

[RJLiberator]

Thanks guys for the help here. Greatly appreciated.