Find B such that AB=BA (Linear Alg)

[QuarkCharmer]

Homework Statement

This is from Lay, 2.1 #11 (the second part). Not a homework problem, just for fun.

Let A be the 3x3 matrix, A = [1,1,1; 1,2,3; 1,4,5]. Find a matrix B such that:
$$AB = BA$$
where B is not the zero or identity matrix

Homework Equations

The Attempt at a Solution

Okay, so I know that typically, AB != BA, since matrix multiplication is non-commutative, but in some cases it can happen. What I did was make some 3x3 matrix B:

B = [a,b,c; d,e,f; g,h,i]

Then I wrote out AB = BA in matrix form, and solved both sides. Let's call AB matrix C, and BA matrix D. I set each entry in C to it's corresponding entry in D to form a system of 9 equations and 9 unknowns. I turned this into a 9x10 augmented matrix and attempted to rref it with my TI-89 but the resulting matrix is too big to display on the screen. I can scroll right and left, but I can't see anything below row 6.

Now, I am pretty confident that, if I did find a value for each entry in B (a,b,c,...,i) then it would form a matrix B such that AB = BA. But I don't think this is the right way to go about this problem at all. I can't believe they would expect me to solve 9 equations with 9 unknowns.

So my questions are:

1.) Would my approach have worked, say, if I computed it in mathematica.

and

2.) What is the proper approach to tackle this problem? I know there must be some way to go about this that is reasonable.Thanks!
-QC

 

[Dustinsfl]

Why not just compute it be hand? You don't need software for a 3x3.

 

[QuarkCharmer]

It's a 9x10 (augmented),
I assumed B was 3x3. A and B both have 9 entries, so that's 9 equations. Unless there is some other way to go about it that I am not seeing.

 

[sponsoredwalk]

If this:

2. Matrix Algebra

Introductory Example: Computer Models in Aircraft Design

2.1 Matrix Operations

2.2 The Inverse of a Matrix

2.3 Characterizations of Invertible Matrices

2.4 Partitioned Matrices

2.5 Matrix Factorizations

...

is the second chapter of your book then read section 2.2 & come back to the question

 

[QuarkCharmer]

That's the book (well the chapters). I guess I'll make note of it and skip it. Seems like they would ask questions possible with only the knowledge provided in chapters 1-2.1.

Would my method have worked, if I cared to work out the 9x9 matrix?

 

[Steely Dan]

Note that if you were to write this matrix equation, it would be three equations, not nine. Therefore the system is under-determined, and there may be many possible solutions. I'm not sure how you got nine equations out of it; they can't be unique.

 

[CornMuffin]

your method would work, and it would find all possible matrices for B, though I certainly wouldn't want to go through all that work

You could just forget about the values of A, and think about what matrices would make AB=BA, you have the zero matrix and the identity, but what other simple matrices can you come up with?

You could also just start off with a 2x2 matrix and see if it gives you some insight.

 

[QuarkCharmer]

I can't think of any way to form this as a 3x3 matrix.

Here is how I get 9x9.

The equations on the right are the simplified equations from the left. Then the standard matrix formed by the 9 equations is at the bottom. In this case vector x has components a,b,c,...


CornMuffin, how would you know to start off with a 2x2?

You could just forget about the values of A, and think about what matrices would make AB=BA, you have the zero matrix and the identity, but what other simple matrices can you come up with?

Well, the identity and zero matrix are the only solutions I can think of for any A, such that AB=BA. Unless they just want some scalar multiple of the identity matrix, which I think would work (trying that now to be sure). Something like [5,0; 0,5], but that's still the identity matrix basically.

 

[Steely Dan]

Unless they just want some scalar multiple of the identity matrix, which I think would work (trying that now to be sure). Something like [5,0; 0,5], but that's still the identity matrix basically.

Sure, but it's not the identity matrix, so it is one possible solution :)

 

[QuarkCharmer]

If that's the answer they were looking for that is pretty lame. There still "could" be some matrix B which satisfies this argument. I'll go with some cheap multiple of the identity matrix for now and revisit after I read the rest of chapter 2.

Thanks for the help

 

[CornMuffin]

Sometimes it is easier to start off with a simpler matrix... by using any simple 2x2 matrix first, it might help you find a matrix B with a larger matrix

edit: there are other matrices that would work that is not a multiple of the identity

 

[Dustinsfl]

Once you get all your equations set equal to each other, you have
$$
\begin{pmatrix}
0&-1&-1&1&0&0&1&0&0&:0\\
1&1&4&0&-1&0&0&-1&0&:0\\
-1&-3&-2&0&0&1&0&0&1&:0\\
1&0&0&1&-1&-1&3&0&0&:0\\
0&1&0&-1&0&-4&0&3&0&:0\\
0&0&0&-1&-3&-3&0&0&3&:0\\
1&0&0&4&0&0&4&-1&-1&:0\\
0&1&0&0&4&0&-1&3&-4&:0\\
0&0&1&0&0&4&-1&-3&0&:0
\end{pmatrix}
$$

 

[Steely Dan]

If that's the answer they were looking for that is pretty lame. There still "could" be some matrix B which satisfies this argument. I'll go with some cheap multiple of the identity matrix for now and revisit after I read the rest of chapter 2.

Thanks for the help

I think that solving nine simultaneous equations is pretty "lame." Finding simple answers to problems that seem complicated is what is cool.

But I don't know what the authors were looking for. That depends on whether they're the type of author that has a sense of humor or not.

 

[CornMuffin]

I think that solving nine simultaneous equations is pretty "lame." Finding simple answers to problems that seem complicated is what is cool.

But I don't know what the authors were looking for. That depends on whether they're the type of author that has a sense of humor or not.

you can use a scalar multiple of the matrix A, or of the inverse of A, or you can even use something like B=A^3+5A+2A^-1 :D

 

[daveb]

I haven't done the math, but perhaps it might be easier to find A^-1 so you get B = A^-1BA.

 

[sponsoredwalk]

There's an extremely easy choice of B, an extremely easy choice...

 

[Bohrok]

If that's the answer they were looking for that is pretty lame. There still "could" be some matrix B which satisfies this argument.

If you make B the inverse of A, that would work; however I don't think that's what the author wants for an answer since inverses are covered in the next section...

 

[Citan Uzuki]

You all are making this much too complicated. Why not just let B=A (I'm guessing this is what you were getting at, sponsoredwalk).

 

[sponsoredwalk]

You all are making this much too complicated. Why not just let B=A (I'm guessing this is what you were getting at, sponsoredwalk).

Maybe A is a bit simpler than A⁵³⁵

 

[I like Serena]

Okay... let's try something.

B is not supposed to be the zero matrix or the identity.
So let's pick the "best" next thing...
Pick a matrix with all entries set to zero, except the top left one...

Yep, that does the trick!

I guess you will need to set more conditions...

Edit: Oh, and yes, B=A or B=A^-1 will also do the trick. :)

 

[I like Serena]

I don't have mathematica either, but I just tried to see what wolframalpha.com makes of it:
http://m.wolframalpha.com/input/?i=solve+{{1%2C1%2C1}%2C{+1%2C2%2C3}%2C{+1%2C4%2C5}}*{{a%2Cb%2Cc}%2C{d%2Ce%2Cf}%2C{g%2Ch%2Ci}}+%3D+{{a%2Cb%2Cc}%2C{d%2Ce%2Cf}%2C{g%2Ch%2Ci}}*{{1%2C1%2C1}%2C{+1%2C2%2C3}%2C{+1%2C4%2C5}}&x=0&y=0

Wolframalpha gives a generic solution based on a, b, and c.
If we pick (1,0,0) we get the identity matrix.
If we pick (1,1,1) we get matrix A back.
If we pick (0,1,1) we get [0,1,1; 1,1,3; 1,4,4].
Take your pick!

 

[Steely Dan]

Okay... let's try something.

B is not supposed to be the zero matrix or the identity.
So let's pick the "best" next thing...
Pick a matrix with all entries set to zero, except the top left one...

Yep, that does the trick!

This doesn't work. Multiplying by $B$ on the right leaves ones on the first column, whereas multiplying by $B$ on the left leaves ones on the first row.

 

[I like Serena]

This doesn't work. Multiplying by $B$ on the right leaves ones on the first column, whereas multiplying by $B$ on the left leaves ones on the first row.

Oops!