# If a matrix commutes with all nxn matrices, then A must be scalar.

1. Feb 8, 2012

### fishshoe

1. The problem statement, all variables and given/known data
Prove: If a matrix A commutes with all matrices B \in M_{nxn}(F), then A must be scalar - i.e., A=diag.(λ,...,λ), for some λ \in F.

2. Relevant equations
If two nxn matrices A and B commute, then AB=BA.

3. The attempt at a solution
I understand that if A is scalar, it will definitely commute with all nxn matrices. But I don't get the intuition behind why commuting with more than one matrix implies that A must be scalar. The way I tried to solve it was by comparing an individual entry in the product, (AB)_{ij} = (BA)_{ij} = (AC)_{ij} = (CA)_{ji}, etc. This implies that
Ʃa_{ik}b_{kj} = Ʃb_{ik}a_{kj} = ...
But I'm not sure how that implies that A is scalar.

2. Feb 8, 2012

### Dick

Think about eigenvectors. Pick special matrices B. Given any vector $v$, extend it to a basis ${v,b_2,b_3,...b_n}$ and define the matrix $B$ by $Bv=v, Bb_k=0$. Can you show that A commuting with B means that v must also be an eigenvector of A?

Last edited: Feb 8, 2012
3. Feb 8, 2012

### fishshoe

I'm trying to figure out what v, b2, b3,..., bn is a basis for. Is it for all nxn matrices?

If Bv=v, then v is an eigenvector of B corresponding to eigenvalue λ=1, and B is the identity operator on the one-dimensional subspace spanned by v.

I know that det(B-I) = 0, so maybe something with determinants?

AB=BA

-> det(AB) = det(BA)

and det(B-I) = 0
det(A)det(B-I)=0
det(A(B-I))=0
det(AB - BI) = 0
det(AB - B) = 0

I'm sorry, that's as far as I've gotten with that. Please let me know if I'm on the right track. Thanks!

4. Feb 8, 2012

### Dick

The vectors are just supposed to be a basis for F^n, the vector space your matrices act on. But, yes, the point is that the eigenvectors of B with eigenvalue 1 are a one dimensional subspace of F^n spanned by v! Now forget about the det's. BAv=ABv put together with Bv=v tells you B(Av)=(Av). So Av is an eigenvector of B with eigenvalue 1. It must lie in the same one dimensional subspace as v. So?

Last edited: Feb 8, 2012
5. Feb 8, 2012

### fishshoe

So since Av is in the same one-dimensional subspace as v, we know that Av is a scalar multiple of v, and so A is a scalar nxn matrix!

But does this apply to any nxn matrix B? Or does it have something to do with the specific B that we defined, such that we have to generalize it further to prove for all cases?

6. Feb 8, 2012

### Dick

No. You don't have to show anything for all matrices B. You can pick any specific ones you want. A has to commute with all of them. What you have so far is that Av is a multiple of v for ANY v. So ANY vector v is an eigenvector of A. So A is a diagonal matrix in any basis. You haven't shown it's a scalar matrix yet. To do that you have to show all of the eigenvectors of A have the same eigenvalue. Keep going.

7. Feb 9, 2012

### fishshoe

So if A is a diagonal matrix in any bases β and γ, then

$[A]_β = diag(a_1,..., a_n)$
and
$[A]_γ = diag(b_1,..., b_n)$

And for the eigenvectors in any basis,

$[A]_βe_i = a_ie_i$

But I'm stuck there. How do I show that

$a_1 = a_2 = ... = a_n$?

8. Feb 9, 2012

### Dick

Suppose A has two linearly independent eigenvectors with two different eigenvalues. We know every vector is an eigenvector of A. See if you can find a contradiction.

Last edited: Feb 9, 2012
9. Feb 9, 2012

### fishshoe

So if I have A = $diag(a_1,...,a_n)$, then

$A\vec{e_1} = a_1\vec{e_1}$
$A\vec{e_2} = a_2\vec{e_2}$
...
$A\vec{e_n} = a_n\vec{e_n}$

But a vector of all 1's should also be an eigenvector of A.

$A * (1,1,...,1)^T = (a_1, a_2, ..., a_n)^T$

And therefore this can only be an eigenvector if all the diagonal elements of A are equal! Is that right?

10. Feb 9, 2012

### Dick

That's right. Put a little more simply, if u and v are eigenvectors with different eigenvalues then u+v can't be an eigenvector.