Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is the set of all $n\times n$ matrices, such that for any fixed matrix B AB=BA, a subspace of the vector space of all $n\times n$ matrices?

Alright, I understand the question and I know what I have to do, basically. I need to show that additive closure and multiplicative closure are satisfied. The problem is, I can't figure out how to do this generally. I tried playing around with $2\times 2$ matrices but that seemed like a dead end. Obviously two such matrices are the 0 matrix and the identity matrix, and those form a subspace, but that doesn't really tell me about all the matrices. Any ideas for how I should be tackling this? I feel like I'm not thinking generally enough.

share|improve this question
    
For a specific B, it might be hard to find any interesting A. To make the exercise easy, just imagine you had found two such guys, A and A'. Is A+A' also one of those guys? Multiply against B and check, just using foil and such. –  Jack Schmidt Oct 20 '11 at 18:39
add comment

3 Answers

up vote 4 down vote accepted

Careful: when you say "multiplicative closure", you have to be clear, since when dealing with $n\times n$ matrices there is a "multiplication" that has nothing to do with the vector space structure (the multiplication of matrices). It is clearer if you refer to it as the "scalar multiplication".

So, fix the matrix $B$. You need to show that:

  1. There is at least one matrix $A$ such that $AB=BA$;
  2. If $A_1$ and $A_2$ are two matrices such that each of them commutes with $B$, then $A_1+A_2$ also commutes with $B$ (closure of your set under vector addition).
  3. If $A$ is a matrix that commutes with $B$, and $k$ is any scalar, then $kA$ also commutes with $B$ (closure of your set under scalar multiplication).

So, with that in mind:

  1. Is there a matrix that necessarily commutes with $B$? (If this thing is really going to be a subspace, it better have what "vector" [i.e., matrix] in it for sure? Try that matrix).

  2. Suppose $A_1$ and $A_2$ both commute with $B$. That is, $A_1B=BA_1$ and $A_2B=BA_2$. You want to show that $(A_1+A_2)$ also commutes with $B$. That is, you want to show that $$(A_1+A_2)B = B(A_1+A_2).$$ Of course, you'll want to use the fact that each of $A_1$ and $A_2$ commutes with $B$, and perhaps some properties you know about matrix multiplication. Is there some property of matrix multiplication that would let you relate $(A_1+A_2)B$ with the products you do know something about, namely $A_1B$ and $A_2B$?

  3. Suppose $A$ commutes with $B$, and $AB=BA$. Let $k$ be a scalar. You want to show that $kA$ also commutes with $B$: that is, you need to prove that $$(kA)B = B(kA).$$ Again, is there some property of matrix multiplication that you know and that might help here?

And if you establish these three, you're done: the set in question is a subspace!

share|improve this answer
    
"There is at least one matrix A such that AB=BA" Thanks, I hadn't remembered that part. Cause a subspace is non-empty. Thanks for the detailed answer, I got it! –  crf Oct 20 '11 at 18:53
    
@Arturo +1-VERY good point about the difference between an actual "multiplication" in the ring-theoretic sense and scalar multiplication in the R-module sense. This is can be a little confusing for the rank amateur. Multiplication in the first sense is an actual operation on the vectors and this can only happen in a vector space if the vector space is itself the field of scalars( i.e. every field is a vector space over itself).,where in the latter case, we simply multiply the vector by a scalar and this results in a magnification of it's length (or norm). –  Mathemagician1234 Oct 20 '11 at 19:11
add comment

Assume you have 2 such matrices, $A_1,A_2$ then $$(A_1+A_2)B = B(A_1+A_2)$$ must hold, so you need to prove this, using the fact that $A_iB=BA_i$ for $i=1,2.$

You then need to show that if $AB=BA$ the it also holds that $(\lambda A)B=B(\lambda A)$.

Tip: Use distribution law for matrices...

share|improve this answer
    
See this is what I need to learn how to do well. Think more generally about matrices rather than about entries. Okay I am going to play around with this, thanks. –  crf Oct 20 '11 at 18:41
    
Yeah, perfect, got it. If I had just thought about it like that for the first like 40 minutes it would have been no problem, hah. Thanks a lot. –  crf Oct 20 '11 at 18:45
add comment

Here is a different kind of answer that is just a lot harder for no reason, but might help you transition.

Suppose we had a specific B, such as: $$B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$ We want to find all A so that AB = BA. Well, we don't know the entries of A, so we replace the unknown numbers with... well unkowns, also known as variables. $$A = \begin{bmatrix} x & y \\ z & t \end{bmatrix}$$

Ok, now we write down what we know about the variables from AB = BA: $$ AB = \begin{bmatrix} x & y \\ z & t \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 2x+4y & 3x+5y \\ 2z+4t & 3z+5t \end{bmatrix} $$ $$ BA = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x & y \\ z & t \end{bmatrix} = \begin{bmatrix} 2x+3z & 2y+3t \\ 4x+5z & 4y+5t \end{bmatrix} $$ For two matrices to be equal, we need to have the entries equal: $$\left\{\begin{align} 2x + 4y &= 2x + 3z \\ 2z+4t &= 4x+5z \\ 4x+5y &= 2y + 3t \\ 3z+5t &= 4y+5t \end{align}\right.$$ What do you know? We have a system of linear equations. We can solve them as usual: find a particular solution (oooo let me let me!): $(x=0,y=0,z=0,t=0)$. Now we know all other solutions are found by adding a "homogeneous solution", which form a subspace.

If we want, we can be even more linear-algebra-y. let's move all the variables to one side:

$$\left\{\begin{align} 0x + 4y - 3z + 0t &= 0 \\ -4x+0y-3z+4t &= 0 \\ 4x+3y +0z -3t &= 0 \\ 0x-4y+3z+0t &= 0 \end{align}\right.$$

We can write it as a matrix equation: $$\left[\begin{array}{rrrr} 0 & 4 & -3 & 0 \\ -4 & 0 & -3 & 4 \\ 4 & 3 & 0 & -3 \\ 0 & -4 & 3 & 0 \end{array}\right] \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

So we are just asking for the null space of the matrix! Clearly that is a subspace. We could even do Gaussian elimination to find which one. I won't.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.