# Math Help - theorem about commuting matrices.

1. ## theorem about commuting matrices.

Hi

I was reading a book on quantum mechanics (thats branch of physics). There is this
theorem which I didn't understand. Now the theorem is about physical observable
quantities like energy,momentum etc.. In quantum mechanics , such physical
observable quantity is represented by an operator. In matrix mechanics , which is a
version of quantum mechanics , such operators are written mathematically as
matrices. So the theorem is basically about matrices. I am going to translate the
theorem in mathematical language , removing words which relate to physics.

Theorem-- If two matrices commute, they possess a common set of eigenvectors.
This is true for both degenerate and non-degenerate eigenvectors.

Now when we have non-degenerate eigenvectors , two such commuting matrices
happen to have common eigenvectors anyway. But I have a question about
degenerate case. In such case , how do we obtain the common eigenvectors ?

2. ## Re: theorem about commuting matrices.

Originally Posted by issacnewton
two such commuting matrices happen to have common eigenvectors anyway.
This may be the important point. Find a copy of a proof and see if it uses the words "if and only if". If so, determine how it is proven in both directions.

3. ## Re: theorem about commuting matrices.

physics books are sloppy when it comes to math "proofs". I don't understand what the author is talking. So I wanted to ask people here. math people are better educated in math than physicists.

4. ## Re: theorem about commuting matrices.

Originally Posted by issacnewton
Hi

I was reading a book on quantum mechanics (thats branch of physics). There is this
theorem which I didn't understand. Now the theorem is about physical observable
quantities like energy,momentum etc.. In quantum mechanics , such physical
observable quantity is represented by an operator. In matrix mechanics , which is a
version of quantum mechanics , such operators are written mathematically as
matrices. So the theorem is basically about matrices. I am going to translate the
theorem in mathematical language , removing words which relate to physics.

Theorem-- If two matrices commute, they possess a common set of eigenvectors.
This is true for both degenerate and non-degenerate eigenvectors.

Now when we have non-degenerate eigenvectors , two such commuting matrices
happen to have common eigenvectors anyway. But I have a question about
degenerate case. In such case , how do we obtain the common eigenvectors ?
we also need the base field $k$ to be algebraically closed. so we have a finite dimensional $k$-vector space $V$ and the operators $T_1,T_2 : V \longrightarrow V$ with $T_1T_2=T_2T_1.$ since $k$ is algebraically closed, $T_2(v)=\lambda v$ for some $\lambda \in k$ and $0 \neq v \in V.$ let

$W=\{x \in V: \ T_2(x)=\lambda x \}.$

then $W$ is a non-zero subspace of $V$ and for any $x \in W$ we have $T_2T_1(x)=T_1T_2(x)=\lambda T_1(x).$ so $T_1(x) \in W.$ thus $T':=T_1|_W$ is an operator on $W.$ let $u \in W$ be any eigenvector of $T'$. clearly $u$ is also an eigenvector of $T_2$ because $u \in W.$ hence $u$ is an eigenvector of both $T_1$ and $T_2$.

5. ## Re: theorem about commuting matrices.

Originally Posted by issacnewton
Hi

I was reading a book on quantum mechanics (thats branch of physics). There is this
theorem which I didn't understand. Now the theorem is about physical observable
quantities like energy,momentum etc.. In quantum mechanics , such physical
observable quantity is represented by an operator. In matrix mechanics , which is a
version of quantum mechanics , such operators are written mathematically as
matrices. So the theorem is basically about matrices. I am going to translate the
theorem in mathematical language , removing words which relate to physics.

Theorem-- If two matrices commute, they possess a common set of eigenvectors.
This is true for both degenerate and non-degenerate eigenvectors.

Now when we have non-degenerate eigenvectors , two such commuting matrices
happen to have common eigenvectors anyway. But I have a question about
degenerate case. In such case , how do we obtain the common eigenvectors ?
Observables in quantum mechanics are always represented by Hermitian operators. The eigenvectors of an Hermitian operator can be chosen to form an orthonormal basis, which implies that they are not degenerate. At least, while you may have degenerate eigenvalues, the geometric multiplicity of degenerate eigenvalues of an Hermitian operator is equal to its algebraic multiplicity. So I'm not entirely sure why you're talking about degenerate eigenvectors. In the degenerate eigenvalue case, you find the eigenvectors in the usual manner. You're not even talking about generalized eigenvectors here.

6. ## Re: theorem about commuting matrices.

Noncomm, thanks for the proof. makes thing clear.

ackbeet , oh yes , i am actually talking about the degenerate eigenvalues ,
not the degenerate eigenvectors..

after checking with the problem i was having , i am able to get the set of
common eigenvectors for the two commuting hermitian matrices having degenerate
eigenvalues. i needed to choose the indeterminate constants properly so that
the set of eigenvectors matched.

but the theorem i quoted doesn't say anything about the hermitian nature of the
matrices. so is it more general theorem , valid for non hermitian commuting matrices as well ?

7. ## Re: theorem about commuting matrices.

Originally Posted by issacnewton
Ackbeet , oh yes , i am actually talking about the degenerate eigenvalues ,
not the degenerate eigenvectors..

after checking with the problem i was having , i am able to get the set of
common eigenvectors for the two commuting hermitian matrices having degenerate
eigenvalues. i needed to choose the indeterminate constants properly so that
the set of eigenvectors matched.

but the theorem i quoted doesn't say anything about the hermitian nature of the
matrices. so is it more general theorem , valid for non hermitian commuting matrices as well ?
According to this wiki, two matrices commute if and only if they are simultaneously diagonalizable. So it appears to be more general than just Hermitian matrices.

8. ## Re: theorem about commuting matrices.

thanks for pointing in the right direction......