How to make commutative 2X2 matrices

[brooksofmaine]

Homework Statement

Create a 2X2 matrix M and a 2X2 matrix N such that MN = NM

Homework Equations

The Attempt at a Solution

 

[Dick]

What's the simplest matrix you can think of?

 

[brooksofmaine]

Dick...thanks for replying. I have to go teach a class. Will respond soon.

 

[brooksofmaine]

Not trying to be flippant, the simplest 2X2 matrix for me would be

0 0
0 0

 

[Dick]

Not trying to be flippant, the simplest 2X2 matrix for me would be

0 0
0 0

That's a great choice. It commutes with any other matrix. Is that good enough?

 

[brooksofmaine]

Not good enough yet because I don't understand the rules behind creating a commutative 2X2 matrix. Here is my humble reasoning so far:

pretend there are two matrices:

a b w x
c d y z

The commutated versions would be

w x a b
y z c d

For them to be commutative then, for example, aw+by (the first step in multiplying the original matrices) would have to equal aw + cx (the first step in multiplying the commutated matrices).

so rule #1 is that aw+by=aw+cx
or simply by=cx.

Rule #1 looks pretty easy to handle.

But when I carry out the second step of matrix multiplication and set the original matrix product equal to the commutated matrix product, I get:

ax+bz = bw+dx

This equation doesn't simplify or help me much.
Neither do the equations for the steps 3 and 4 of the multiplication process.

So my approach to finding a rules for making commutative matrices looks as if it's going nowhere.

Is there a better approach?

 

[Dick]

You want to find the conditions on a matrix that will allow it to commute with all other matrices? Take a general matrix [[a,b],[c,d]] and see what the conditions are that it commute with [[1,0],[0,0]] and [[0,1],[0,0]]. (The inner brackets in my matrix notation represent the rows of the matrix).

 

[HallsofIvy]

You do understand, don't you that there is no one solution? Given any matrix, there exist an infinite number of matrices that commute with it.

Let's just say that the first matrix is
$$\left[\begin{array}{cc}1 & 4 \\ 5 & 2\end{array}\right]$$
(for no particular reason- I just made up that matrix)

Multiplying that on both left and right by
$$\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$$
gives
$$\left[\begin{array}{cc}a+ 4c & 4a+ 2b \\ c+ 5d & 4c+ 2d\end{array}\right]= \left[\begin{array}{cc}a+ 4c & b+ 4d \\ 5a+ 2c & 5b+ 2d\end{array}\right]$$

which gives the four equations a+ 5b= a+ 4c, 4a+ 2b= b+ 4d, c+ 5d= 5a+ 2c, and 4c+ 2d= 5b+ 2d. Those equations, of course, are NOT independent. Notice that both the first and fourth equations reduce to the same thing: 5b= 4c. As I said before there exist an infinite number of matrices which will commute with the one given.

We can also reduce 4a+ 2b= b+ 4d to 4a+ b= 4d and the third equation, c+ 5d= 5a+ 2c to 5d= 5a + c. From 5b= 4c, b= 4c/5. Then the second equation becomes 4a+ 4c/5= 4d or a= d- c/5. Putting a= d- c/5 into the last equation, 5d= 5d- c+ c which is automatically satisfied. Essentially that means that I can choose c and d to be anything I want and then solve for a and b.

If I take c= 5, d= 1, I get a= 1- 5/5= 0, b= 4(5)/5= 4. I will leave it to you to show that
$$\left[\begin{array}{cc}1 & 4 \\5 & 2\end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ 5 & 1\end{array}\left]= \left[\begin{array}{cc} 0 & 4 \\ 5 & 1\end{array}\left]\left[\begin{array}{cc}1 & 4 \\5 & 2\end{array}\right]$$.

Or I can take c= 10 (obviously, choosing c a multiple of 5 avoids fractions), d= -2, we have that a= -2-2= -4 and b= 8
$$\left[\begin{array}{cc}1 & 4 \\5 & 2\end{array}\right]\left[\begin{array}{cc} -4 & 8 \\ 10 & -2\end{array}\left]= \left[\begin{array}{cc} -4 & 8 \\ 10 & -2\end{array}\left]\left[\begin{array}{cc}1 & 4 \\5 & 2\end{array}\right]$$.

 

[HallsofIvy]

A commutative matrix? One that commutes with all other matrices?

Bother! I just went to a lot of work for nothing!. I will say that Dick's suggestion is very good. If you have a matrix that will commute with every matrix, then it will have to commute with
$$\left[\begin{array}{cc}1 & 0 \\ 0 & 0\end{array}\right]$$
$$\left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right]$$
$$\left[\begin{array}{cc}0 & 0 \\ 1 & 0\end{array}\right]$$
and
$$\left[\begin{array}{cc}0 & 0 \\ 0 & 1\end{array}\right]$$
the "standard basis" 2 by 2 matrices. You should find a fairly obvious answer.

 

[brooksofmaine]

Wow...you have given me much to look over and contemplate. I have some classes to teach now so I will reflect on your thoughts later. Thanks very much.

 

[HallsofIvy]

I'm kind of hoping that class you have to teach isn't Linear Algebra!:rofl:

 

[brooksofmaine]

I like the humor! The class is music composition, and linear algebra is new to me, but I'm fascinated by it. Maybe I'll work it in somehow. Meanwhile, I will study your posts.

 

[Dick]

So your hobby is linear algebra?? The summary (omitting details) of the previous posts is that the question of exactly which matrices commute depends on the details of the nature of the specific matrices. As Halls showed, it's not that hard to find matrices that commute with a given matrix. I was trying to show you how to show that the only matrices that commute with ALL other matrices are multiples of the identity.

 

[brooksofmaine]

My son Ian and I reviewed everything you so kindly wrote, and all is clear. If you ever need some commutative 2X2 matrices, just let me know and I'll prepare a few. Thanks for the truly helpful help.