Subspace and basis problem

1. Nov 12, 2008

DWill

1. The problem statement, all variables and given/known data
Let B be a fixed n x n matrix, and let X_B = { A e M_n so that AB = BA }. In other words, X_B is the set of all matrices which commute with B.

(a) Prove that X_B is a subspace of M_n.
(b) Let B =
[
1 0
2 -1
]

Find a basis for X_B and write its dimension.

(c) Is your basis in part (b) orthogonal? If not, make it orthogonal using the
Gram-Schmidt algorithm. Then, normalize the basis to make it orthonor-
mal.

2. Relevant equations
A e M_n ---> A belongs to set M_n

3. The attempt at a solution
For part (a) what is M_n ? Is this some general notation for a certain matrix (like I for identity)? I think I need to know that before I can try proving anything. Thanks

2. Nov 12, 2008

Staff: Mentor

M_n has to represent n x n matrices. For a given n x n matrix B, for AB to be defined, A has to have n columns. For the same B, for BA to be defined, A has to have n rows.

IOW, A has to be n x n for both products to be defined.

3. Nov 13, 2008

DWill

Ok so I asked my teacher about part (a) and she said that for it to be a subspace, I have to check the following 4 properties:

2. Closed under scalar multiplication
3. Contains 0
4. Closed under inverses

For property 1, she said that take A and C in X_B, then if (A+C)B = B(A+C) then this shows A+C is also in X_B. I don't understand this, why does showing the above equation is true show that it is closed under addition? I think I can show properties 2 and 3, but property 4 I'm also confused by. Can someone show me how to prove if A is in X_B then -A is also in X_B?

One more question, in part (b) I am given the matrix B. To find a basis, I need to find the matrices that commute with B, and then find the set of matrices that span X_B and are linearly independent? Any hints are helpful.

Thanks again

4. Nov 13, 2008

Staff: Mentor

I think your teacher might be new at this subject. To show that a subset U of a vector space V is actually a subspace of V, all you need to show are:
1. if x and y are in U then x + y is in U.
2. if x is in U and c is a scalar, then cx is in U.
You can even cut the two steps above to one step by showing that if x and y are in U, and c is a scalar, then c(x + y) is in U.
The four steps your teacher gave are covered by the two I gave above.
You are correct to not understand this, because it is not showing that the set is closed under addition; it is showing that addition in set is commutative. It also lessens my confidence in your teacher's knowledge.
For the property 1 that she gave, you need to show that if x is in U and y is in U, then x + y is in U. Period. Using your variables, if A is in the set (I called this set U) and B is in the set, then A + B is in the set.

The second condition that I showed that must be satisfied is that if c is a scalar and x is in U, then cx is in U. Consider the scalar -1.

The matrix B that you were given is a 2 x 2 matrix with one entry 0. Any matrix A has to be 2 x 2 as well, for reasons I gave earlier. Just put in values for A (a, b, c, and d, for example) and multiply both products AB and BA. The system of equations will describe the subspace you're looking for.

5. Nov 13, 2008

Office_Shredder

Staff Emeritus
I think you might be new at this subject. For a subset U of a vector space V to be a subspace, it needs to be non-empty

No, you're wrong here. The definition of XB is that it's the set of matrices that commute with B. So showing that if A commutes with B and C commutes with B, then A+C commutes with B is exactly showing that the set is closed under addition.

6. Nov 13, 2008

Staff: Mentor

Shredder,
You got me. Of course the set under consideration has to be nonempty. As for the rest, it slipped my mind that the set under consideration was the set of matrices that commute with B.
Mark

7. Nov 13, 2008

DWill

Ok thanks for the help above.

So like Mark said for (b) I let A = (a, b, c, d) and found AB and BA and set them equal. I end up getting the following system of equations:

a + 2b = a
-b = b
c + 2d = 2a - c
-d = 2b - d

I'm not sure if I did this correctly but I found this to be the basis (all 2x2 matrices):
{ (0, 0, 1, 1) , (1, 0, 0, -1), (1, 0, -1, 0) }

Assuming this is correct (please point out any mistakes!), to know whether or not this is an orthogonal basis do I just see if the dot product between each pair of matrices is 0 or not?

8. Nov 14, 2008

HallsofIvy

Staff Emeritus
so b= 0 and the first equation is just "a= a".

Since b= 0, the last equation is just -d= -d. The third equation is 2c+ 2d= a2a and dividing by 2, a= c+ d. Taking c= 1, d= 0, a= 1 so (1, 0, 1, 0). Taking c= 0, d= 1, a= 1 so (1, 0, 0, 1). (1, 0, 0, -1), that is, a= 1, b= 0, c= -1, d= 0 does NOT satisfy c+ 2d= 2a- c. Since every thing reduces to two equations, b= 0 and a= c+ d, this subspace is 4- 2= two dimensional and has a basis consisting of two matrices, not three.

Yes. What is the "dot product" of matrices?